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Maths M.C.Q (1 Marks)

Trigonometric Identities · CBSE STD 10 (CBSE - English Medium). 89 questions.

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M.C.Q (1 Marks)

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89 questions · auto-graded multiple-choice test.

MCQ 11 Mark
$\frac{1+\tan^2\text{A}}{1+\cot^2\text{A}}$ is equal to:
  • A
    $\sec^2\text{A}$
  • B
    $-1$
  • C
    $\cot^2\text{A}$
  • ✓
    $\tan^2\text{A}$
Answer
Correct option: D.
$\tan^2\text{A}$
$\frac{1+\tan^2\text{A}}{1+\cot^2\text{A}}=\frac{\sec^2\text{A}}{\text{cosec}^2\text{A}}=\frac{\sin^2\text{A}}{\cos^2\text{A}}$
$=\tan^2\text{A}$
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MCQ 21 Mark
$2(\sin^6\theta+\cos^6\theta)-3(\sin^4\theta+\cos^4\theta)$ is equal to:
  • A
    0
  • B
    1
  • ✓
    -1
  • D
    None of these.
Answer
Correct option: C.
-1
$2(\sin^6\theta+\cos^6\theta)-3(\sin^4\theta+\cos^4\theta)$
$=2\big[(\sin^2\theta)^3+(\cos^2\theta)^3\big]-3\big[(\sin^2\theta)^2+(\cos^2\theta)^2\big]$
$=2\big[(\sin^2\theta+\cos^2\theta)(\sin^4\theta+\cos^4\theta-\sin^2\theta\cos^2\theta)\big]$
$=-3\big[(\sin^2\theta+\cos^2\theta)^2-2\sin^2\theta\cos^2\theta\big]$
$\{\because\ \text{a}^3+\text{b}^2=(\text{a}+\text{b})^3-3\text{ab}(\text{a}+\text{b})\}$
$=2\big[1(\sin^2\theta)^2+(\cos^2\theta)^2+2\sin^2\theta\cos^2\theta-3\sin^2\theta+\cos^2\theta\big]$
$=-3\big[(1)^2-2\sin^2\theta\cos^2\theta\big]$
$=2\big[(\sin^2\theta+\cos^2\theta)^2-3\sin^2\theta\cos^2\theta\big]-3\big[1-2\sin^2\theta\cos^2\theta\big]$
$=2\big[1-3\sin^2\theta\cos^2\theta\big]-3\big[1-2\sin^2\theta\cos^2\theta\big]$
$=2-6\sin^2\theta\cos^3\theta-3+6\sin^2\theta\cos^2\theta$
$=-1$
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MCQ 31 Mark
If $\text{a}\cos\theta+\text{b}\sin\theta=4\text{ and a}\sin\theta-\text{b}\cos\theta=3,$ then $a^2+b^2=0$
  • A
    $7$
  • B
    $12$
  • ✓
    $25$
  • D
    None of these.
Answer
Correct option: C.
$25$
Given,
$\text{a}\cos\theta+\text{b}\sin\theta=4,$
$\text{a}\sin\theta-\text{b}\cos\theta=3$
Squaring and then adding the above two equations, we have
$(\text{a}\cos\theta+\text{b}\sin\theta)^2+(\text{a}\sin\theta-\text{b}\cos\theta)^2=(4)^2+(3)^2$
$\Rightarrow (\text{a}^2\cos^2\theta+\text{b}^2\sin^2\theta)+2\times\text{a}\cos\theta\times\sin\theta)$
$=(\text{a}^2\sin^2\theta+\text{b}^2\cos^2\theta-2\text{a}\sin\theta\times\text{b}\cos\theta)=16+9$
$\Rightarrow\ \text{a}^2\cos^2\theta+\text{b}^2\sin^2\theta+2\text{ab}\sin\theta\cos\theta$
$=\text{a}^2\sin^2\theta+\text{b}^2\cos^2\theta-2\text{ab}\sin\theta\cos\theta=25$
$\Rightarrow \text{a}^2\cos^2\theta+\text{b}^2\sin^2\theta+\text{a}^2\sin^2\theta+\text{b}^2\cos^2\theta=25$
$\Rightarrow ( \text{a}^2\cos^2\theta+\text{a}^2\sin^2\theta)+(\text{b}^2\sin^2\theta+\text{b}^2\cos^2\theta)=25$
$\Rightarrow \text{a}^2(\cos^2\theta+\sin^2\theta)+\text{b}^2(\sin^2\theta+\cos^2\theta)=25$
$\Rightarrow \text{a}^2(1)+\text{b}^2(1)=25$
$\Rightarrow \text{a}^2+\text{b}^2=25$
Hence, the correct option is $(C).$
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MCQ 41 Mark
If $\text{x}=\text{r}\sin\theta\cos\phi,\text{y}=\text{r}\sin\phi$ and ${z}=\text{r}\cos\theta,$ then:
  • ✓
    $x^2+y^2+z^2=r^2$
  • B
    $x^2+y^2-z^2=r^2$
  • C
    $x^2-y^2+z^2=r^2$
  • D
    $z^2+y^2-x^2=r^2$
Answer
Correct option: A.
$x^2+y^2+z^2=r^2$
$\text{x}=\text{r}\sin\theta\cos\phi$
$\text{y}=\text{r}\sin\theta\sin\phi$
$\text{z}=\text{r}\cos\theta$
Squaring and adding these equations, we get
$\text{x}^2+\text{y}^2+\text{z}^2=(\text{r}\sin\theta\cos\phi)^2+(\text{r}\sin\theta\sin\phi)^2+(\text{r}\cos\theta)^2$
$\Rightarrow\ \text{x}^2+\text{y}^2+\text{z}^2=\text{r}^2\sin^2\theta\cos^2\phi+\text{r}^2\sin^2\theta\sin^2\phi+\text{r}^2\cos^2\theta$
$\Rightarrow\ \text{x}^2+\text{y}^2+\text{z}^2=(\text{r}^2\sin^2\theta\cos^2\phi+\text{r}^2\sin^2\theta\sin^2\phi)+\text{r}^2\cos^2\theta$
$\Rightarrow\text{x}^2+\text{y}^2+\text{z}^2=\text{r}^2\sin^2\theta(\cos^2\phi+\sin^2\phi)+\text{r}^2\cos^2\theta$
$\Rightarrow\ \text{x}^2+\text{y}^2+\text{z}^2=\text{r}^2\sin^2\theta(1)+\text{r}^2\cos^2\theta$
$\Rightarrow\ \text{x}^2+\text{y}^2+\text{z}^2=\text{r}^2\sin^2\theta+\text{r}^2\cos^2\theta$
$\Rightarrow\ \text{x}^2+\text{y}^2+\text{z}^2=\text{r}^2(\sin^2\theta+\cos^2\theta)$
$\Rightarrow\ \text{x}^2+\text{y}^2+\text{z}^2=\text{r}^2(1)$
$\Rightarrow\ \text{x}^2+\text{y}^2+\text{z}^2=\text{r}^2$
Hence, the correct option is $(A).$
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MCQ 51 Mark
$(\sec\text{A}+\tan\text{A})(1-\sin\text{A})=$
  • A
    $\sec\text{A}$
  • B
    $\sin\text{A}$
  • C
    $\text{cosec A}$
  • ✓
    $\cos\text{A}$
Answer
Correct option: D.
$\cos\text{A}$
The given expression is $(\sec\text{A}+\tan\text{A})(1-\sin\text{A})$
Simplifying the given expression, we have
$(\sec\text{A}+\tan\text{A})(1-\sin\text{A})$
$=\Big(\frac{1}{\cos\text{A}}+\frac{\sin\text{A}}{\cos\text{A}}\Big)(1-\sin\text{A})$
$=\Big(\frac{1+\sin\text{A}}{\cos\text{A}}\Big)\times(1-\sin\text{A})$
$=\frac{(1+\sin\text{A})(1-\sin\text{A})}{\cos\text{A}}$
$=\frac{1-\sin^2\text{A}}{\cos\text{A}}$
$=\frac{\cos^2\text{A}}{\cos\text{A}}$
$=\cos\text{A}$
Therefore, the correct option is (d).
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MCQ 61 Mark
$\frac{\tan\theta}{\sec\theta-1}+\frac{\tan\theta}{\sec\theta + 1}$ is equal to:
  • A
    $2\tan\theta$
  • B
    $2\sec\theta$
  • ✓
    $2\text{cosec }\theta$
  • D
    $2\tan\theta\sec\theta$
Answer
Correct option: C.
$2\text{cosec }\theta$
The givne expression is $\frac{\tan\theta}{\sec\theta-1}+\frac{\tan\theta}{\sec\theta+1}$
Simplifying the given expression, we have
$\frac{\tan\theta}{\sec\theta-1}+\frac{\tan\theta}{\sec\theta+1}$
$=\frac{\tan\theta(\sec\theta+1)+\tan\theta(\sec\theta-1)}{(\sec\theta-1)(\sec\theta+1)}$
$=\frac{\tan\theta\sec\theta+\tan\theta+\tan\theta\sec\theta-\tan\theta}{\sec^2\theta-1}$
$=\frac{2\tan\theta\sec\theta}{\tan^2\theta}$
$=\frac{2\sec\theta}{\tan\theta}$
$=\frac{2\frac{1}{\cos\theta}}{\frac{\sin\theta}{\cos\theta}}$
$=2\frac{1}{\sin\theta}$
$=2\text{cosec }\theta$
Therefore, the correct option is (c).
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MCQ 71 Mark
If $\cos(\alpha+\beta)=0,$ then $\sin(\alpha-\beta)$ can be reduced to:
  • A
    $\cos\beta$
  • ✓
    $\cos2\beta$
  • C
    $\sin\alpha$
  • D
    $\sin2\alpha$
Answer
Correct option: B.
$\cos2\beta$
$\cos(\alpha+\beta)=0$
$\Rightarrow\ \alpha+\beta=90^\circ \big[\because\ \cos90^\circ=0\big]$
$\Rightarrow\ \alpha=90^\circ-\beta\ .....(\text{i})$
$\sin(\alpha-\beta)=\sin(90^\circ-\beta-\beta)\ \big[\text{using (i)}\big]$
$=\sin\big(90^\circ-2\beta\big)$
$=\cos2\beta\ \big[\because \sin(90^\circ -\theta)=\cos\theta\big]$
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MCQ 81 Mark
If $\text{a}\cos\theta+\text{b}\sin\theta=\text{m}$ and $\text{a}\sin\theta-\text{b}\cos\theta=\text{n},$ then $a^2 + b^2 =$
  • A
    $m^2-n^2$
  • B
    $m^2 n^2$
  • C
    $n^2-m^2$
  • ✓
    $m^2+n^2$
Answer
Correct option: D.
$m^2+n^2$
$\text{a}\cos\theta+\text{b}\sin\theta=\text{m}$
$\text{a}\sin\theta-\text{b}\cos\theta=\text{n}$
Squaring and adding
$\text{a}^2\cos^2\theta+\text{b}^2\sin^2\theta+2\text{ab }\sin\theta\cos\theta=\text{m}^2$
$\text{a}^2\sin^2\theta+\text{b}^2\cos^2\theta-2\text{ab }\sin\theta\cos\theta=\text{n}^2$
$\text{a}^2(\cos^2\theta+\sin^2\theta)+\text{b}^2(\sin^2\theta+\cos^2\theta)$
$=\text{m}^2+\text{n}^2\ \{\sin^2\theta+\cos^2\theta=1\}$
$\Rightarrow \text{a}^2+1+\text{b}^2\times1=\text{m}^2-\text{n}^2$
$\Rightarrow \text{a}^2+\text{b}^2=\text{m}^2+\text{n}^2$
Hence, $a^2+b^2=m^2+n^2$.
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MCQ 91 Mark
$\frac{\sin\theta}{1-\cot\theta}+\frac{\cos\theta}{1-\tan\theta}$ is equal to:
  • A
    $0$
  • B
    $1$
  • ✓
    $\sin\theta+\cos\theta$
  • D
    $\sin\theta-\cos\theta$
Answer
Correct option: C.
$\sin\theta+\cos\theta$
$\frac{\sin\theta}{1-\cot\theta}+\frac{\cos\theta}{1-\tan\theta}=\frac{\sin\theta}{1-\frac{\cos\theta}{\sin\theta}}+\frac{\cos\theta}{1-\frac{\sin\theta}{\cos\theta}}$
$=\frac{\sin\theta\times\sin\theta}{\sin\theta-\cos\theta}+\frac{\cos\theta\times\cos\theta}{\cos\theta-\sin\theta}$
$=\frac{\sin^2\theta}{\sin\theta-\cos\theta}-\frac{\cos^2\theta}{\sin\theta-\cos\theta}$
$=\frac{\sin^2\theta-\cos^2\theta}{\sin\theta-\cos\theta}$
$=\frac{(\sin\theta+\cos\theta)(\sin\theta-\cos\theta)}{\sin\theta-\cos\theta}$
$=\sin\theta+\cos\theta$
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MCQ 101 Mark
If $\text{x}=\text{a}\sec\theta\cos\phi,\text{y}=\text{b}\sec\theta\sin\phi\text{ and z}=\text{c}\tan\theta,$ then $\frac{\text{x}^2}{\text{a}^2}+\frac{\text{y}^2}{\text{b}^2}=$
  • A
    $\frac{\text{z}^2}{\text{c}^2}$
  • B
    $1-\frac{\text{z}^2}{\text{c}^2}$
  • C
    $\frac{\text{z}^2}{\text{c}^2}-1$
  • ✓
    $1+\frac{\text{z}^2}{\text{c}^2}$
Answer
Correct option: D.
$1+\frac{\text{z}^2}{\text{c}^2}$
$\text{x}=\text{a}\sec\theta\cos\phi$
$\text{y}=\text{b}\sec\theta\sin\phi$
$\text{z}=\text{c}\tan\theta$
$\frac{\text{x}}{\text{a}}=\sec\theta\cos\phi\ .....(\text{i})$
$\frac{\text{y}}{\text{b}}=\sec\theta\sin\phi\ .....(\text{ii})$
$\frac{\text{z}}{\text{c}}=\tan\theta\ .....(\text{iii})$
Squaring and adding (i) and (ii)
$\frac{\text{x}^2}{\text{a}^2}+\frac{\text{y}^2}{\text{b}^2}=\sec^2\theta\cos^2\phi+\sec^2\theta\sin^2\phi$
$=\sec^2\theta(\cos^2\phi+\sin^2\phi)$
$=\sec^2\theta\times1=\sec^2\theta$
Squaring (iii) and subtracting from (iv)
$\frac{\text{x}^2}{\text{a}^2}+\frac{\text{y}^2}{\text{b}^2}-\frac{\text{z}^2}{\text{c}^2}=\sec^2\theta-\tan^2\theta=1$
$\frac{\text{x}^2}{\text{a}^2}+\frac{\text{y}^2}{\text{b}^2}=1+\frac{\text{z}^2}{\text{c}^2}$
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MCQ 111 Mark
The value of $\sqrt{\frac{1+\cos\theta}{1-\cos\theta}}$ is:
  • A
    $\cot\theta-\text{cosec }\theta$
  • ✓
    $\text{cosec }\theta+\cot\theta$
  • C
    $\text{cosec}^2\theta+\cot^2\theta$
  • D
    $(\cot\theta+\text{cosec }\theta)^2$
Answer
Correct option: B.
$\text{cosec }\theta+\cot\theta$
The given expression is $\sqrt{\frac{1+\cos\theta}{1-\cos\theta}}$
Multiplying both the numerator and denominator under the root by $(1+\cos\theta)$, we have
$\sqrt{\frac{(1+\cos\theta)(1+\cos\theta)}{(1+\cos\theta)(1-\cos\theta)}}$
$=\sqrt{\frac{(1+\cos\theta)^2}{(1-\cos^2\theta)}}$
$=\sqrt{\frac{(1+\cos\theta)^2}{\sin^2\theta}}$
$=\frac{1+\cos\theta}{\sin\theta}$
$=\frac{1}{\sin\theta}+\frac{\cos\theta}{\sin\theta}$
$=\text{cosec }\theta+\cot\theta$
Therefore, the correcr choise is (b).
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MCQ 121 Mark
$\frac{\sin\theta}{1+\cos\theta}$ is equal to:
  • A
    $\frac{1+\cos\theta}{\sin\theta}$
  • B
    $\frac{1-\cos\theta}{\cos\theta}$
  • ✓
    $\frac{1-\cos\theta}{\sin\theta}$
  • D
    $\frac{1-\sin\theta}{\cos\theta}$
Answer
Correct option: C.
$\frac{1-\cos\theta}{\sin\theta}$
The given expression is $\frac{\sin\theta}{1+\cos\theta}$
Multiplying both the numerator and denominator under the root by $(1-\cos\theta)$, we have
$\frac{\sin\theta}{1+\cos\theta}$
$=\frac{\sin\theta(1-\cos\theta)}{(1+\cos\theta)(1-\cos\theta)}$
$=\frac{\sin\theta(1-\cos\theta)}{1-\cos^2\theta}$
$=\frac{\sin\theta(1-\cos\theta)}{\sin^2\theta}$
$=\frac{1-\cos\theta}{\sin\theta}$
Therefore, the correct option is (C).
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MCQ 131 Mark
$\frac{\cot\theta}{\cot\theta-\cot3\theta}+\frac{\tan\theta}{\tan\theta-\tan3\theta}$ is equal to:
  • A
    0
  • ✓
    1
  • C
    -1
  • D
    2
Answer
Correct option: B.
1
$=\frac{\cot\theta}{\cot\theta-\cot3\theta}+\frac{\tan\theta}{\tan\theta-\tan3\theta}$
$=\frac{\cot\theta\tan\theta-\cot\theta\tan3\theta+\cot\theta\tan\theta-\tan\theta\cot3\theta}{(\cot\theta-\cot3\theta)(\tan\theta-\tan3\theta)}$
$\{\tan\theta\cot\theta=1\}$
$\Rightarrow\ \frac{1-\cot\theta\tan3\theta+1-\tan\theta\cot3\theta}{\cot\theta\tan\theta-\cot\theta\tan3\theta-\tan\theta\cot3\theta+\cot3\theta\tan3\theta}$
$=\frac{2-\cot\theta\tan3\theta-\tan\theta\cot3\theta}{1-\cot\theta\tan3\theta-\tan\theta\cot3\theta+1}$
$=\frac{2-\cot\theta\tan3\theta-\tan\theta\cot3\theta}{2-\cot\theta\tan\theta-\tan\theta\cot3\theta}=1$
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MCQ 141 Mark
$9\sec^2\text{A}-9\tan^2\text{A}$ is equal to:
  • A
    1
  • ✓
    9
  • C
    8
  • D
    0
Answer
Correct option: B.
9
Given,
$9\sec^2\text{A}-9\tan^2\text{A}$
$=9(\sec^2\text{A}-\tan^2\text{A})$
We know that, $\sec^2-\tan^2\text{A}=1$
Therefore, $9\sec^2\text{A}-9\tan^2\text{A}=9$
Hence, the correct option is (b).
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MCQ 151 Mark
If $\cos9\theta=\sin\theta\text{ and }9\theta<90^\circ,$ then value of $\tan6\theta$ is:
  • A
    $\frac{1}{\sqrt{3}}$
  • ✓
    $\sqrt{3}$
  • C
    $1$
  • D
    $0$
Answer
Correct option: B.
$\sqrt{3}$
$\cos(9\theta)=\sin\theta$
$\Rightarrow\ \sin(90^\circ-9\theta)=\sin\theta$
$\Rightarrow\ 90^\circ-9\theta=\theta$
$\Rightarrow\ 90^\circ=\theta+9\theta$
$\Rightarrow\ \theta=10$
$\tan6\theta=\tan6$
$=\tan60^\circ=\sqrt{3}$
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MCQ 161 Mark
The value of $\sin(45^\circ+\theta)-\cos(45^\circ-\theta)$ is equal to:
  • A
    $2\cos\theta$
  • ✓
    $0$
  • C
    $2\sin\theta$
  • D
    $1$
Answer
Correct option: B.
$0$
We know that, $\sin(90-\theta)=\cos\theta$
So,
$\sin(45^\circ+\theta)=\cos\big[90-(45^\circ+\theta)\big]=\cos(45^\circ-\theta)$
$\therefore\ \sin(45^\circ+\theta)-\cos(45^\circ-\theta)$
$= \cos(45^\circ-\theta)-\cos(45^\circ-\theta)$
$=0$
Hence, the correct answer is option (b).
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MCQ 171 Mark
$\cos^4\text{A}-\sin^4\text{A}$ is equal to:
  • A
    $2\cos^2\text{A}+1$
  • ✓
    $2\cos^2\text{A}-1$
  • C
    $2\sin^2\text{A}-1$
  • D
    $2\sin^2\text{A}+1$
Answer
Correct option: B.
$2\cos^2\text{A}-1$
$\cos^4\text{A}-\sin^4\text{A}=(\cos^2\text{A}+\sin^2\text{A})(\cos^2\text{A}-\sin^2\text{A})$
$=1(\cos^2\text{A}-\sin^2\text{A})=\cos^2\text{A}-(1-\cos^2\text{A})$
$=\cos^2\text{A}-1+\cos^2\text{A}$
$=2\cos^2\text{A}-1$
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MCQ 181 Mark
If $\sec\theta+\tan\theta=\text{x},$ then $\sec\theta=$
  • A
    $\frac{\text{x}^2+1}{\text{x}}$
  • ✓
    $\frac{\text{x}^2+1}{2\text{x}}$
  • C
    $\frac{\text{x}^2-1}{2\text{x}}$
  • D
    $\frac{\text{x}^2-1}{\text{x}}$
Answer
Correct option: B.
$\frac{\text{x}^2+1}{2\text{x}}$
Given, $\sec\theta+\tan\theta=\text{x}$
We know that,
$\sec^2\theta-\tan^2\theta=1$
$\Rightarrow (\sec\theta+\tan\theta)(\sec\theta-\tan\theta)=1$
$\Rightarrow\ \text{x}(\sec\theta-\tan\theta)=\frac{1}{\text{x}}$
Now,
$\sec\theta+\tan\theta=\text{x},$
$\sec\theta-\tan\theta=\frac{1}{\text{x}}$
Adding the two equations, we get
$(\sec\theta+\tan\theta)+(\sec\theta-\tan\theta)=\text{x}+\frac{1}{\text{x}}$
$\Rightarrow\ \sec\theta+\tan\theta+\sec\theta-\tan\theta=\frac{\text{x}^2+1}{\text{x}}$
$\Rightarrow\ 2\sec\theta=\frac{\text{x}^2+1}{\text{x}}$
$\Rightarrow\ \sec\theta=\frac{\text{x}^2+1}{2\text{x}}$
Therefore, the correct choice is (b).
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MCQ 191 Mark
If $\sin\theta+\sin^2\theta=1$ then $\cos^2\theta+\cos^4\theta$:
  • A
    -1
  • ✓
    1
  • C
    0
  • D
    None of these.
Answer
Correct option: B.
1
$\sin\theta+\sin^2\theta=1$
$\Rightarrow\ \sin\theta=1-\sin^2\theta$
$\Rightarrow\ \sin\theta=\cos^2\theta$
$\cos^2\theta+\cos^4\theta=\sin\theta+\sin^2\theta\ \{\because \cos^2\theta=\sin\theta\}$
$\Rightarrow\ \cos^2\theta+\cos^4\theta=1$
$\{\because \sin\theta+\sin^2\theta=1(\text{given})\}$
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MCQ 201 Mark
The value of $(1+\cot\theta-\text{cosec }\theta)(1+\tan\theta+\sec\theta)$ is
  • A
    1
  • ✓
    2
  • C
    4
  • D
    0
Answer
Correct option: B.
2
$(1+\cot\theta-\text{cosec }\theta)(1+\tan\theta+\sec\theta)$
$=\Big(1+\frac{\cos\theta}{\sin\theta}-\frac{1}{\sin\theta}\Big)\Big(1+\frac{\sin\theta}{\cos\theta}+\frac{1}{\cos\theta}\Big)$
$=\frac{(\sin\theta+\cos\theta-1)(\cos\theta+\sin\theta+1)}{\sin\theta\times\cos\theta}$
$=\frac{\{(\sin\theta+\cos\theta)-1\}\{(\cos\theta+\sin\theta)+1\}}{\sin\theta\cos\theta}$
$=\frac{(\cos\theta+\sin\theta)^2-1}{\sin\theta\cos\theta}$
$=\frac{\cos^2\theta+\sin^2\theta+2\sin\theta\cos\theta-1}{\sin\theta\cos\theta}$
$=\frac{1+2\sin\theta\cos\theta-1}{\sin\theta\cos\theta}=\frac{2\sin\theta\cos\theta}{\sin\theta\cos\theta}$
$=2$
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MCQ 211 Mark
If $\text{a}\cot\theta+\text{b}\text{ cosec}\theta=\text{p and b}\cot\theta+\text{a cosec }\theta=\text{q},$ then $p^2 - q^2 =$
  • A
    $a^2-b^2$
  • ✓
    $b^2-a^2$
  • C
    $a^2+b^2$
  • D
    $b-a$
Answer
Correct option: B.
$b^2-a^2$
$\text{a}\cot\theta+\text{b cosec }\theta=\text{p}$
$\text{b}\cot\theta+\text{a cosec }\theta=\text{q}$
Squaring and subtracting,
$\text{p}^2-\text{q}^2=(\text{a}\cot\theta+\text{b cosec }\theta)^2-(\text{b}\cot\theta+\text{a cosec}\theta)^2$
$=\text{a}^2\cot^2\theta+\text{b}^2\text{cosec}^2\theta+2\text{ab}\cot\theta\text{ cosec }\theta$
$=-(\text{b}^2\cot^2\theta+\text{a}^2\text{cosec}^2\theta+2\text{ab}\cot\theta\text{ cosec }\theta)$
$=\text{a}^2\cot^2\theta+\text{b}^2\text{cosec}^2\theta+2\text{ab}\cot\theta\text{ cosec }\theta$
$=-\text{b}^2\cot^2\theta-\text{a}^2\text{cosec}^2\theta-2\text{ab}\cot\theta\text{ cosec }\theta$
$=\text{a}^2(\cot^2\theta-\text{cosec}^2\theta)+\text{b}^2(\text{cosec}^2\theta-\cot^2\theta)$
$=-\text{a}^2(\text{cosec}^2\theta-\cot^2\theta)+\text{b}^2(\text{cosec}^2\theta-\cot^2\theta)$
$=-\text{a}^2\times1+\text{b}^2\times1$
$=\text{b}^2-\text{a}^2$
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MCQ 221 Mark
$(\text{cosec }\theta-\sin\theta)(\sec\theta-\cos\theta)(\tan\theta+\cot\theta)$ is equal to:
  • A
    0
  • ✓
    1
  • C
    -1
  • D
    None of these.
Answer
Correct option: B.
1
$(\text{cosec }\theta-\sin\theta)(\sec\theta-\cos\theta)(\tan\theta+\cot\theta)$
$=\Big(\frac{1}{\sin\theta}-\sin\theta\Big)\Big(\frac{1}{\cos\theta}-\cos\theta\Big)$
$\Big(\frac{\sin\theta}{\cos\theta}+\frac{\cos\theta}{\sin\theta}\Big)$
$=\frac{1-\sin^2\theta}{\sin\theta}\times\frac{1-\cos^2\theta}{\cos\theta}\times\frac{\sin^2\theta+\cos^2\theta}{\sin\theta\cos\theta}$
$=\frac{\cos^2\theta}{\sin\theta}\times\frac{\sin^2\theta}{\cos\theta}\times\frac{1}{\sin\theta\cos\theta}$
$=\frac{\sin^2\theta\cos^2\theta}{\sin^2\theta\cos^2\theta}=1$
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MCQ 231 Mark
$\sqrt{\frac{1+\sin\theta}{1-\sin\theta}}$ is equal to:
  • ✓
    $\sec\theta+\tan\theta$
  • B
    $\sec\theta-\tan\theta$
  • C
    $\sec^2\theta+\tan^2\theta$
  • D
    $\sec^2\theta-\tan^2\theta$
Answer
Correct option: A.
$\sec\theta+\tan\theta$
$\sqrt{\frac{1+\sin\theta}{1-\sin\theta}}=\sqrt{\frac{(1+\sin\theta)(1+\sin\theta)}{(1-\sin\theta)(1+\sin\theta)}}$
$=\sqrt{\frac{(1+\sin\theta)^2}{1-\sin^2\theta}}$
$=\sqrt{\frac{(1+\sin\theta)^2}{\cos^2\theta}}=\frac{1+\sin\theta}{\cos\theta}=\frac{1}{\cos\theta}+\frac{\sin\theta}{\cos\theta}$
$=\sec\theta+\tan\theta$
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MCQ 241 Mark
$\sec^4\text{A}-\sec^2\text{A}$ is equal to:
  • A
    $\tan^2\text{A}-\tan^4\text{A}$
  • B
    $\tan^4\text{A}-\tan^2\text{A}$
  • ✓
    $\tan^4\text{A}+\tan^2\text{A}$
  • D
    $\tan^2\text{A}+\tan^4\text{A}$
Answer
Correct option: C.
$\tan^4\text{A}+\tan^2\text{A}$
$\sec^4-\sec^2\text{A}=\sec^2\text{A}(\sec^2\text{A}-1)$
$=(1+\tan^2\text{A})\tan^2\text{A}$
$\begin{cases}\sec^2\text{A}=1+\tan^2\text{A} \\ \sec^2\text{A}-1=\tan^2\text{A}\end{cases}$
$=\tan^2\text{A}+\tan^4\text{A}$
$=\tan^4\text{A}+\tan^2\text{A}$
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MCQ 251 Mark
If $\sec\theta+\tan\theta=\text{x}\sec\theta+\tan\theta=\text{x},$ then $\tan\theta=\tan\theta=$
  • A
    $\frac{\text{x}^2+1}{\text{x}}$
  • B
    $\frac{\text{x}^2-1}{\text{x}}$
  • C
    $\frac{\text{x}^2+1}{2\text{x}}$
  • ✓
    $\frac{\text{x}^2-1}{2\text{x}}$
Answer
Correct option: D.
$\frac{\text{x}^2-1}{2\text{x}}$
$\sec\theta+\tan\theta=\text{x}\ .....\text{(i)}$
We know that
$\sec^2\theta-\tan^2\theta=1$
$\Rightarrow \ (\sec\theta+\tan\theta)(\sec\theta-\tan\theta)=1$
$\Rightarrow\ \text{x}(\sec\theta-\tan\theta)=1$
$\Rightarrow\ \sec\theta-\tan\theta=\frac{1}{\text{x}}\ .....(\text{ii})$
Subtracting (ii) from (i)
$2\tan\theta=\text{x}-\frac{1}{\text{x}}=\frac{\text{x}^2-1}{\text{x}}$
$\tan\theta=\frac{\text{x}^2-1}{2\text{x}}$
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MCQ 261 Mark
If $\triangle\text{ABC}$ is right angled at C, then the value of $\cos(\text{A}+\text{B})$ is:
  • ✓
    $0$
  • B
    $1$
  • C
    $\frac{1}{2}$
  • D
    $\frac{\sqrt{3}}{2}$
Answer
Correct option: A.
$0$
In a right angled traingle ABC, $\triangle\text{C}$ is a righta angle.
We know that, the sum of angles of a triangle is 180º.
$\therefore\ \angle\text{A}+\angle\text{B}+\angle\text{C}=180^\circ$
$\Rightarrow\ \angle\text{A}+\angle\text{B}+90^\circ=180^\circ$
$\Rightarrow\ \angle\text{A}+\angle\text{B}=90^\circ$
$\therefore\ \cos(\text{A}+\text{B})=\cos90^\circ=0$
Hence, the correct answer is option (a).
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MCQ 271 Mark
If $\text{a}\cos\theta-\text{b}\sin\theta=\text{c},$ then $\text{a}\sin\theta+\text{b}\cos\theta=$
  • A
    $\pm\sqrt{\text{a}^2+\text{b}^2+\text{c}^2}$
  • ✓
    $\pm\sqrt{\text{a}^2+\text{b}^2-\text{c}^2}$
  • C
    $\pm\sqrt{\text{c}^2-\text{a}^2-\text{b}^2}$
  • D
    None of these.
Answer
Correct option: B.
$\pm\sqrt{\text{a}^2+\text{b}^2-\text{c}^2}$
$\text{a}\cos\theta-\text{b}\sin\theta=\text{c}$
Squaring,
$\text{a}^2\cos^2\theta+\text{b}^2\sin^2\theta-2\text{ab}\sin\theta\cos\theta=\text{c}^2$
$\Rightarrow\ \text{a}^2(1-\sin^2\theta)+\text{b}^2(1-\cos^2\theta)-2\text{ab}\sin\theta\cos\theta=\text{c}^2$
$\Rightarrow\ \text{a}^2-\text{a}^2\sin^2\theta+\text{b}^2(1-\cos^2\theta)-2\text{ab}\sin\theta\cos\theta=\text{c}^2$
$\Rightarrow\ -\text{a}^2\sin^2\theta-\text{b}^2\cos^2\theta-2\text{ab}\sin\theta\cos\theta=\text{c}^2-\text{a}^2-\text{b}^2$
$\Rightarrow\ \text{a}^2\sin^2\theta+\text{b}^2\cos\theta+2\text{ab}\sin\theta\cos\theta=\text{a}^2+\text{b}^2-\text{c}^2 $ $\ (\text{Dividing by}-1)$
$(\text{a}\sin\theta+\text{b}\cos\theta)^2=\text{a}^2+\text{b}^2-\text{c}^2$
$\therefore\ \text{a}\sin\theta+\text{b}\cos\theta=\pm\sqrt{\text{a}^2+\text{b}^2-\text{c}^2}$
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MCQ 281 Mark
If $x = a \cos \theta$ and $y = b \sin \theta$, then $b ^2 x ^2+ a ^2 y ^2=$
 
  • ✓
    $a^2 b^2$
  • B
    $ab$
  • C
    $a^4 b^4$
  • D
    $a^2+b^2$
Answer
Correct option: A.
$a^2 b^2$
$\text{x}=\text{a}\cos\theta,\text{y}=\text{b}\sin\theta\ .....(\text{i})$
$\text{bx}=\text{ab}\cos\theta,\text{ay}=\text{ab}\sin\theta\ .....(\text{ii})$
Adding $(i)$ and $(ii)$ we get,
$=\text{b}^2\text{x}^2+\text{a}^2\text{b}^2\cos^2\theta+\text{a}^2\text{b}^2\sin^2\theta$
$=\text{a}^2\text{b}^2(\cos^2\theta+\sin^2\theta)$
$=\text{a}^2\text{b}^2\times1$
$=\text{a}^2\text{b}^2$
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MCQ 291 Mark
If $x=a \sec \theta$ and $y=b \tan \theta$, then $b^2 x^2-a^2 y^2=$
  • A
    $ab$
  • B
    $a^2-b^2$
  • C
    $a^2+b^2$
  • ✓
    $a^2 b^2$
     
Answer
Correct option: D.
$a^2 b^2$
 
Given, $\text{x}=\text{a}\sec\theta,\text{y}=\text{b}\tan\theta$
So,
$\text{b}^2\text{y}^2-\text{a}^2\text{y}^2$
$=\text{b}^2(\text{a}\sec\theta)^2-\text{a}^2(\text{b}\tan\theta)^2$
$=\text{b}^2\text{a}^2\sec^2\theta-\text{a}^2\text{b}^2\tan^2\theta$
$=\text{b}^2\text{a}^2(\sec^2\theta-\tan^2\theta)$
We know that,
$\sec^2\theta-\tan^2\theta=1$
$\text{b}^2\text{x}^2-\text{a}^2\text{y}^2=\text{a}^2\text{b}^2$
Hence, the correct option is $(D).$
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MCQ 301 Mark
The value of $\sin^229^\circ+\sin^261^\circ$ is:
  • ✓
    $1$
  • B
    $0$
  • C
    $2\sin^2{29}^\circ$
  • D
    $2\cos^2{61}^\circ$
Answer
Correct option: A.
$1$
$\sin^2{29}^\circ+\sin^2{61}^\circ=\sin^2{29}^\circ+\sin^2{(90^\circ-29^\circ})$
$=\sin^2{29}+\cos^2{29}^\circ$
$(\sin^2\theta+\cos^2\theta=1)$
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MCQ 311 Mark
$(1+\tan\theta+\sec\theta)(1+\cot\theta-\text{cosec }\theta)=$
  • A
    0
  • B
    1
  • ✓
    2
  • D
    -1
Answer
Correct option: C.
2
$(1+\tan\theta+\sec\theta)(1+\cot\theta-\text{cosec }\theta)$
$=1+\cot\theta-\text{cosec }\theta+\tan\theta+\cot\theta\tan\theta$
$=-\tan\theta\text{ cosec }\theta+\sec\theta+\sec\theta\cot\theta-\sec\theta\text{ cosec }\theta$
$=1+\frac{\cos\theta}{\sin\theta}-\frac{1}{\sin\theta}+\frac{\sin\theta}{\cos\theta}+1-\frac{\sin\theta}{\cos\theta}\times\frac{1}{\sin\theta}$
$=\frac{1}{\cos\theta}+\frac{1}{\cos\theta}\times\frac{\cos\theta}{\sin\theta}-\frac{1}{\cos\theta}\times\frac{1}{\sin\theta}$
$=2+\frac{\cos\theta}{\sin\theta}+\frac{\sin\theta}{\cos\theta}-\frac{1}{\sin\theta}-\frac{1}{\cos\theta}$
$=\frac{1}{\cos\theta}+\frac{1}{\sin\theta}-\frac{1}{\sin\theta\cos\theta}$
$=2+\frac{\sin\theta}{\cos\theta}+\frac{\cos\theta}{\sin\theta}-\frac{1}{\sin\theta\cos\theta}$
$=2+\frac{\sin^2\theta+\cos^2\theta}{\sin\theta\cos\theta}-\frac{1}{\sin\theta\cos\theta}$
$=2+\frac{1}{\sin\theta\cos\theta}-\frac{1}{\sin\theta\cos\theta}=2$
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MCQ 321 Mark
If $\cos\text{A}+\cos^2\text{A}=1,$ then $\sin^2\text{A}+\sin^4\text{A}=$
  • A
    -1
  • B
    0
  • ✓
    1
  • D
    None of these.
Answer
Correct option: C.
1
Given,
$\cos\text{A}+\cos^2\text{A}=1$
$\Rightarrow\ 1-\cos^2\text{A}=\cos\text{A}$
So,
$\sin^2\text{A}+\sin^4\text{A}$
$=\sin^2\text{A}+\sin^2\text{A}\sin^2\text{A}$
$=\sin^2\text{A}+(1-\cos^2\text{A})(1-\cos^2\text{A})$
$=\sin^2\text{A}+\cos\text{A}\cos\text{A}$
$=\sin^2\text{A}+\cos^2\text{A}$
$=1$
Hence, the correct option is (c).
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MCQ 331 Mark
If $\sin\theta-\cos\theta=0,$ then the value of $\sin^4\theta+\cos^4\theta$ is:
  • A
    $1$
  • B
    $\frac{3}{4}$
  • ✓
    $\frac{1}{2}$
  • D
    $\frac{1}{4}$
Answer
Correct option: C.
$\frac{1}{2}$
$\sin\theta-\cos\theta=0$
$\Rightarrow\ \sin\theta=\cos\theta$
$\Rightarrow\ \frac{\sin\theta}{\cos\theta}=1$
$\Rightarrow\ \tan\theta=1$
$\Rightarrow\ \theta=45^\circ$
Now, put the value of $\theta$ in the given equation
$\sin^4\theta+\cos^4\theta$
$=\sin^4{45}^\circ+\cos^4{45}^\circ$
$=\Big(\frac{1}{\sqrt{2}}\Big)^4+\Big(\frac{1}{\sqrt{2}}\Big)^4$
$=\frac{1}{4}+\frac{1}{4}=\frac{2}{4}$
$=\frac{1}{2}$
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MCQ 341 Mark
$\frac{1+\tan ^2 A}{1+\cot ^2 A}$ is equal to
  • A
    $\sec ^2 A$
  • B
    -1
  • C
    $\cot ^2 A$
  • ✓
    $\tan ^2 A$
Answer
Correct option: D.
$\tan ^2 A$
D
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MCQ 351 Mark
$(\sec A+\tan A)(1-\sin A)=$
  • A
    $\sec A$
  • B
    $\sin A$
  • C
    $\operatorname{cosec} A$
  • ✓
    $\cos A$
Answer
Correct option: D.
$\cos A$
D
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MCQ 371 Mark
The value of $(\sec A+\tan A)(1-\sin A)$ is
  • A
    $\sec A$
  • B
    $\sin A$
  • C
    $\operatorname{cosec} A$
  • ✓
    $\cos A$
Answer
Correct option: D.
$\cos A$
D
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MCQ 381 Mark
$3 \sin ^2 \theta+4 \cos ^2 \theta$ is cqual to
  • A
    1
  • B
    2
  • C
    $\sin ^2 \theta+3$
  • ✓
    $\cos ^2 \theta+3$
Answer
Correct option: D.
$\cos ^2 \theta+3$
(D)$\cos ^2 \theta+3$
$3 \sin ^2 \theta+4 \cos ^2 \theta=\left(3 \sin ^2 \theta+3 \cos ^2 \theta\right)+\cos ^2 \theta$
$=3\left(\sin ^2 \theta+\cos ^2 \theta\right)+\cos ^2 \theta=3+\cos ^2 \theta$
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MCQ 391 Mark
If $\sin \alpha=\frac{\sqrt{3}}{2}, \cos \beta=\frac{\sqrt{3}}{2}$, then $\tan \alpha \tan \beta$ is
  • A
    $\sqrt{3}$
  • B
    $\frac{1}{\sqrt{3}}$
  • ✓
    1
  • D
    $0$
Answer
Correct option: C.
1
C
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MCQ 401 Mark
If $\frac{x}{3}=2 \sin A, \frac{y}{3}=2 \cos A$, then the value of $x^2+y^2$ is
  • ✓
    36
  • B
    9
  • C
    6
  • D
    18
Answer
Correct option: A.
36
A
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MCQ 411 Mark
If $\cos (\alpha+\beta)=0$, then the value of $\cos \left(\frac{\alpha+\beta}{2}\right)$ is equal to
  • ✓
    $\frac{1}{\sqrt{2}}$
  • B
    $\frac{1}{2}$
  • C
    $0$
  • D
    $\sqrt{2}$
Answer
Correct option: A.
$\frac{1}{\sqrt{2}}$
A
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MCQ 421 Mark
If $\sec \theta-\tan \theta=m$, then the value of $\sec \theta+\tan \theta$ is
  • A
    $1-\frac{1}{m}$
  • B
    $m^2-1$
  • ✓
    $\frac{1}{m}$
  • D
    $-m$
Answer
Correct option: C.
$\frac{1}{m}$
C
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MCQ 431 Mark
If 0 is an acute angle of a right angled triangle, then which of the following equation is not true?
  • A
    $\sin \theta \cot \theta=\cos \theta$
  • B
    $\cos \theta \tan \theta=\sin \theta$
  • C
    $\operatorname{cosec}^2 \theta-\cot ^2 \theta=1$
  • ✓
    $\tan ^2 \theta-\sec ^2 \theta=1$
Answer
Correct option: D.
$\tan ^2 \theta-\sec ^2 \theta=1$
(D)$\tan ^2 \theta-\sec ^2 \theta=1$
$\sin \theta \cot \theta=\sin \theta \times \frac{\cos \theta}{\sin \theta}=\cos \theta$. So, option (a) is true.
$\cos \theta \tan \theta=\cos \theta \times \frac{\sin \theta}{\cos \theta}=\sin \theta$. So, option (b) is true.
$1+\cot ^2 \theta=\operatorname{cosec}^2 \theta$ and $1+\tan ^2 \theta=\sec ^2 \theta \Rightarrow \operatorname{cosec}^2 \theta-\cot ^2 \theta=1$ and $\sec ^2 \theta-\tan ^2 \theta=1$ So, option (c) is true but option (d) is not true.
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MCQ 441 Mark
$\frac{\cos ^2 \theta}{\sin ^2 \theta}-\frac{1}{\sin ^2 \theta}$, in simplified form, is
  • A
    $\tan ^2 \theta$
  • B
    $\sec ^2 \theta$
  • C
    1
  • ✓
    -1
Answer
Correct option: D.
-1
(D)-1
$\frac{\cos ^2 \theta}{\sin ^2 \theta}-\frac{1}{\sin ^2 \theta}$
$=\left(\frac{\cos \theta}{\sin \theta}\right)^2-\left(\frac{1}{\sin \theta}\right)^2$
$=\cot ^2 \theta-\operatorname{cosec}^2 \theta$
$=-\left(\operatorname{cosec}^2 \theta-\cot ^2 \theta\right)=-1$
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MCQ 451 Mark
If $\sin \theta-\cos \theta=0$, then the walue of $\sin ^6 \theta+\cos ^6 \theta$ is
  • A
    $\frac{2}{3}$
  • B
    $\frac{1}{3}$
  • C
    $\frac{3}{4}$
  • ✓
    $\frac{1}{4}$
Answer
Correct option: D.
$\frac{1}{4}$
(D)$\frac{1}{4}$
We have,
$
\begin{array}{l}
\sin \theta-\cos \theta=0 \Rightarrow \sin \theta=\cos \theta \Rightarrow \tan \theta=1 \Rightarrow \theta=45^{\circ} \\
\sin ^6 \theta+\cos ^6 \theta=(\sin \theta)^6+(\cos \theta)^6=\left(\sin 45^{\circ}\right)^6+\left(\cos 45^{\circ}\right)^6=\left(\frac{1}{\sqrt{2}}\right)^6+\left(\frac{1}{\sqrt{2}}\right)^6=\frac{1}{8}+\frac{1}{8}=\frac{1}{4}
\end{array}
$
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MCQ 461 Mark
If $(\sin \theta+\operatorname{cosec} \theta)^2+(\cos \theta+\sec \theta)^2=k+\tan ^2 \theta+\cot ^2 \theta$, then $k=$
  • A
    1
  • B
    2
  • C
    5
  • ✓
    7
Answer
Correct option: D.
7
(D)7
We find that
$
(\sin \theta+\operatorname{cosec} \theta)^2+(\cos \theta+\sec \theta)^2
$
$
\begin{array}{l}
=\sin ^2 \theta+2 \sin \theta \operatorname{cosec} \theta+\operatorname{cosec}^2 \theta+\cos ^2 \theta+2 \cos \theta \sec \theta+\sec ^2 \theta \\
=\sin ^2 \theta+2+1+\cot ^2 \theta+\cos ^2 \theta+2+1+\tan ^2 \theta \\
=\left(\sin ^2 \theta+\cos ^2 \theta\right)+6+\cot ^2 \theta+\tan ^2 \theta=7+\tan ^2 \theta+\cot ^2 \theta
\end{array}
$
Hence, $k=7$.
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MCQ 471 Mark
If $k+1=\sec ^2 \theta(1+\sin \theta)(1-\sin \theta)$, then $k=$
  • A
    1
  • B
    2
  • ✓
    $0$
  • D
    -1
Answer
Correct option: C.
$0$
(C)0
We have,
$
\begin{array}{l}
k+1=\sec ^2 \theta(1+\sin \theta)(1-\sin \theta) \\
\Rightarrow \quad k+1=\frac{1}{\cos ^2 \theta} \times\left(1-\sin ^2 \theta\right) \Rightarrow k+1=\frac{\cos ^2 \theta}{\cos ^2 \theta} \Rightarrow k+1=1 \Rightarrow k=0
\end{array}
$
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MCQ 481 Mark
$(\operatorname{cosec} \theta-\sin \theta)(\sec \theta-\cos \theta)(\tan \theta+\cot \theta)$ is equal to
  • A
    $2 \sqrt{2}$
  • B
    $0$
  • ✓
    1
  • D
    $\sqrt{2}$
Answer
Correct option: C.
1
(C)1
$
\begin{aligned}
\text { } & (\operatorname{cosec} \theta-\sin \theta)(\sec \theta-\cos \theta)(\tan \theta+\cot \theta) \\
= & \left(\frac{1}{\sin \theta}-\sin \theta\right)\left(\frac{1}{\cos \theta}-\cos \theta\right)\left(\frac{\sin \theta}{\cos \theta}+\frac{\cos \theta}{\sin \theta}\right) \\
= & \frac{1-\sin ^2 \theta}{\sin \theta} \times \frac{1-\cos ^2 \theta}{\cos \theta} \times \frac{\sin ^2 \theta+\cos ^2 \theta}{\sin \theta \cos \theta}=\frac{\cos ^2 \theta}{\sin \theta} \times \frac{\sin ^2 \theta}{\cos \theta} \times \frac{1}{\sin \theta \cos \theta}=1
\end{aligned}
$
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MCQ 491 Mark
If $1+\sin ^2 \theta=3 \sin \theta \cos \theta$, then $\tan \theta$ can take values
  • ✓
    $1, \frac{1}{2}$
  • B
    1,2
  • C
    $\frac{1}{2}, 2$
  • D
    none of these
Answer
Correct option: A.
$1, \frac{1}{2}$
(A) $1, \frac{1}{2}$
We have, $1+\sin ^2 \theta=3 \sin \theta \cos \theta$
Dividing both sides by $\cos ^2 \theta$, we get
$
\begin{aligned}
& \frac{1}{\cos ^2 \theta}+\frac{\sin ^2 \theta}{\cos ^2 \theta}=\frac{3 \sin \theta \cos \theta}{\cos ^2 \theta} \\
\Rightarrow & \sec ^2 \theta+\tan ^2 \theta=3 \tan \theta \\
\Rightarrow & 1+\tan ^2 \theta+\tan ^2 \theta=3 \tan \theta \Rightarrow 2 \tan ^2 \theta-3 \tan \theta+1=0 \Rightarrow(2 \tan \theta-1)(\tan \theta-1)=0 \Rightarrow \tan \theta=1, \frac{1}{2}
\end{aligned}
$
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MCQ 501 Mark
If $\sqrt{3} \cot ^2 \theta-4 \cot \theta+\sqrt{3}=0$, then the value of $3\left(\cot ^2 \theta+\tan ^2 \theta\right)$ is
  • A
    15
  • B
    7
  • ✓
    10
  • D
    4
Answer
Correct option: C.
10
(C)10
We have,
$
\begin{array}{ll}
& \sqrt{3} \cot ^2 \theta-4 \cot \theta+\sqrt{3}=0 \\
\Rightarrow & \sqrt{3} \cot ^2 \theta-3 \cot \theta-\cot \theta+\sqrt{3}=0 \\
\Rightarrow & \sqrt{3} \cot \theta(\cot \theta-\sqrt{3})-(\cot \theta-\sqrt{3})=0 \\
\Rightarrow & (\sqrt{3} \cot \theta-1)(\cot \theta-\sqrt{3})=0
\end{array}
$
$\Rightarrow \quad \sqrt{3} \cot \theta-1=0$ or, $\cot \theta-\sqrt{3}=0 \Rightarrow \cot \theta=\frac{1}{\sqrt{3}}$ or, $\cot \theta=\sqrt{3} \Rightarrow \theta=60^{\circ}$ or $\theta=30^{\circ}$
If $\theta=60^{\circ}$, then $3\left(\cot ^2 \theta+\tan ^2 \theta\right)=3\left(\cot ^2 60^{\circ}+\tan ^2 60^{\circ}\right)=3\left(\frac{1}{3}+3\right)=10$
If $\theta=30^{\circ}$, then $3\left(\cot ^2 \theta+\tan ^2 \theta\right)=3\left(\cot ^2 30^{\circ}+\tan ^2 30^{\circ}\right)=3\left(\frac{1}{3}+3\right)=10$
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MCQ 511 Mark
If $15 \tan ^2 \theta+4 \sec ^2 \theta=23$, then the value of $(\sec \theta+\operatorname{cosec} \theta)^2-\sin ^2 \theta$ is
  • A
    7
  • ✓
    $\frac{15}{2}$
  • C
    $\frac{9}{2}$
  • D
    $\frac{11}{2}$
Answer
Correct option: B.
$\frac{15}{2}$
(B) $\frac{15}{2}$
We have,
$15 \tan ^2 \theta+4 \sec ^2 \theta=23$
$\Rightarrow \quad 15 \tan ^2 \theta+4\left(1+\tan ^2 \theta\right)$$=23 \Rightarrow 19 \tan ^2 \theta=19$
$\Rightarrow \tan ^2 \theta=1 \Rightarrow \tan \theta=1 \Rightarrow \theta=45^{\circ}$
$\therefore \quad(\sec \theta+\operatorname{cosec} \theta)^2-\sin ^2 \theta$
$=\left(\sec 45^{\circ}+\operatorname{cosec} 45^{\circ}\right)^2-\sin ^2 45^{\circ}$
$=(\sqrt{2}+\sqrt{2})^2-\left(\frac{1}{\sqrt{2}}\right)^2$
$=8-\frac{1}{2}=\frac{15}{2}$
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MCQ 521 Mark
If $\sin \theta+\cos \theta=\sqrt{2}$, then $\tan \theta+\cot \theta=$
  • A
    1
  • ✓
    2
  • C
    3
  • D
    4
Answer
Correct option: B.
2
(B)2
$
\begin{array}{l}
\text { We have, } \sin \theta+\cos \theta=\sqrt{2} \\
\Rightarrow \quad(\sin \theta+\cos \theta)^2=2 \\
\Rightarrow \quad \sin ^2 \theta+\cos ^2 \theta+2 \sin \theta \cos \theta=2 \Rightarrow 1+2 \sin \theta \cos \theta=2 \Rightarrow 2 \sin \theta \cos \theta=1 \Rightarrow \sin \theta \cos \theta=\frac{1}{2} \\
\therefore \quad \tan \theta+\cot \theta=\frac{\sin \theta}{\cos \theta}+\frac{\cos \theta}{\sin \theta}=\frac{\sin ^2 \theta+\cos ^2 \theta}{\sin \theta \cos \theta}=\frac{1}{\sin \theta \cos \theta}=2
\end{array}
$
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MCQ 531 Mark
$\frac{\tan ^2 \theta}{1+\tan ^2 \theta}+\frac{\cot ^2 \theta}{1+\cot ^2 \theta}=$
  • ✓
    1
  • B
    $2 \tan ^2 \theta$
  • C
    $2 \cot ^2 \theta$
  • D
    $2 \sec ^2 \theta$
Answer
Correct option: A.
1
(A) 1
$\frac{\tan ^2 \theta}{1+\tan ^2 \theta}+\frac{\cot ^2 \theta}{1+\cot ^2 \theta}$
$=\frac{\tan ^2 \theta}{\sec ^2 \theta}+\frac{\cot ^2 \theta}{\operatorname{cosec}^2 \theta}$
$=\tan ^2 \theta \times \frac{1}{\sec ^2 \theta}+\cot ^2 \theta \times \frac{1}{\operatorname{cosec}^2 \theta}$
$=\frac{\sin ^2 \theta}{\cos ^2 \theta} \times \cos ^2 \theta+\frac{\cos ^2 \theta}{\sin ^2 \theta} \times \sin ^2 \theta$
$=\sin ^2 \theta+\cos ^2 \theta=1$
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MCQ 541 Mark
$\tan ^4 \theta+\tan ^2 \theta=$
  • A
    $\sec ^2 \theta-2 \sec ^4 \theta$
  • B
    $2 \sec ^2 \theta-2 \sec ^4 \theta$
  • C
    $\sec ^2 \theta-\sec ^4 \theta$
  • ✓
    $\sec ^4 \theta-\sec ^2 \theta$
Answer
Correct option: D.
$\sec ^4 \theta-\sec ^2 \theta$
(D) $\sec ^4 \theta-\sec ^2 \theta \tan ^4 \theta+\tan ^2 \theta$
$=\tan ^2 \theta\left(\tan ^2 \theta+1\right)$
$=\tan ^2 \theta \sec ^2 \theta=\left(\sec ^2 \theta-1\right) \sec ^2 \theta$
$=\sec ^4 \theta-\sec ^2 \theta$.
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MCQ 551 Mark
If $\cos A=\frac{3}{5}$, then the value $9+9 \tan ^2 A$ is
  • A
    9
  • B
    16
  • ✓
    25
  • D
    34
Answer
Correct option: C.
25
(C)25
$
\text {} 9+9 \tan ^2 A=9\left(1+\tan ^2 A\right)=9 \sec ^2 A=\frac{9}{\cos ^2 A}=9 \times\left(\frac{5}{3}\right)^2=25
$
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MCQ 561 Mark
$(1+\tan A-\sec A)(1+\tan A+\sec A)=$
  • ✓
    $2 \tan A$
  • B
    $2 \sin A$
  • C
    $2 \sec A$
  • D
    $2 \cot A$
Answer
Correct option: A.
$2 \tan A$
(A)$2 \tan A$
$
\begin{aligned}
\text {} & (1+\tan A-\sec A)(1+\tan A+\sec A) \\
= & (1+\tan A)^2-\sec ^2 A=1+\tan ^2 A+2 \tan A-\sec ^2 A=\sec ^2 A+2 \tan A-\sec ^2 A=2 \tan
\end{aligned}
$
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MCQ 571 Mark
The value of $\left(1+\tan ^2 \theta\right)(1+\sin \theta)(1-\sin \theta)(1+\cos \theta)(1-\cos \theta)\left(1+\cot ^2 \theta\right)$ is
  • ✓
    1
  • B
    -1
  • C
    $0$
  • D
    -2
Answer
Correct option: A.
1
(A)1
$\begin{array}{l}\left(1+\tan ^2 \theta\right)(1+\sin \theta)(1-\sin \theta)(1+\cos \theta)(1-\cos \theta)\left(1+\cot ^2 \theta\right) \\ =\sec ^2 \theta\left(1-\sin ^2 \theta\right)\left(1-\cos ^2 \theta\right) \operatorname{cosec}^2 \theta=\frac{1}{\cos ^2 \theta} \times \cos ^2 \theta \times \sin ^2 \theta \times \frac{1}{\sin ^2 \theta}=1\end{array}$
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MCQ 581 Mark
The value of $\cos ^2 \theta+\frac{1}{1+\cot ^2 \theta}$ is
  • A
    $0$
  • ✓
    1
  • C
    2
  • D
    -1
Answer
Correct option: B.
1
(B)1
$\cos ^2 \theta+\frac{1}{1+\cot ^2 \theta}=\cos ^2 \theta+\frac{1}{\operatorname{cosec}^2 \theta}=\cos ^2 \theta+\sin ^2 \theta=1$
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MCQ 601 Mark
If $a \cos \theta+b \sin \theta=m$ and $a \sin \theta-b \cos \theta=n$, then $a^2+b^2=$
  • A
    $m^2-n^2$
  • B
    $m^2 n^2$
  • C
    $n^2-m^2$
  • ✓
    $m^2+n^2$
Answer
Correct option: D.
$m^2+n^2$
D
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MCQ 611 Mark
If $\sin \theta+\sin ^2 \theta=1$, then $\cos ^2 \theta+\cos ^4 \theta=$
  • A
    -1
  • ✓
    1
  • C
    $0$
  • D
    none of these
Answer
Correct option: B.
1
B
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MCQ 621 Mark
If $x=r \sin \theta \cos \phi, y=r \sin \theta \sin \phi$ and $z=r \cos \theta$, then
  • ✓
    $x^2+y^2+z^2=r^2$
  • B
    $x^2+y^2-z^2=r^2$
  • C
    $x^2-y^2+z^2=r^2$
  • D
    $z^2+y^2-x^2=r^2$
Answer
Correct option: A.
$x^2+y^2+z^2=r^2$
A
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MCQ 631 Mark
If $a \cot \theta+b \operatorname{cosec} \theta=p$ and $b \cot \theta+a \operatorname{cosec} \theta=q$, then $p^2-q^2=$
  • A
    $a^2-b^2$
  • ✓
    $b^2-a^2$
  • C
    $a^2+b^2$
  • D
    $b-a$
Answer
Correct option: B.
$b^2-a^2$
B
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MCQ 641 Mark
If $a \cos \theta+b \sin \theta=4$ and $a \sin \theta-b \cos \theta=3$, then $a^2+b^2=$
  • A
    7
  • B
    12
  • ✓
    25
  • D
    none of these
Answer
Correct option: C.
25
C
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MCQ 651 Mark
$2\left(\sin ^6 \theta+\cos ^6 \theta\right)-3\left(\sin ^4 \theta+\cos ^4 \theta\right)$ is equal to
  • A
    $0$
  • B
    1
  • ✓
    -1
  • D
    none of these
Answer
Correct option: C.
-1
C
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MCQ 661 Mark
$\frac{\cot \theta}{\cot \theta-\cot 3 \theta}+\frac{\tan \theta}{\tan \theta-\tan 3 \theta}$ is equal to
  • A
    $0$
  • ✓
    1
  • C
    -1
  • D
    2
Answer
Correct option: B.
1
B
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MCQ 671 Mark
If $1+\sin ^2 \alpha=3 \sin \alpha \cos \alpha$, then the values of $\cot \alpha$ are
  • A
    $-1,1$
  • B
    0,1
  • ✓
    1,2
  • D
    $-1,-1$
Answer
Correct option: C.
1,2
C
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MCQ 681 Mark
If $\cos (\alpha+\beta)=0$, then $\sin (\alpha-\beta)$ can be reduced to
  • A
    $\cos \beta$
  • ✓
    $\cos 2 \beta$
  • C
    $\sin \alpha$
  • D
    $\sin 2 \alpha$
Answer
Correct option: B.
$\cos 2 \beta$
B
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MCQ 691 Mark
If $a \cos \theta-b \sin \theta=c$, then $a \sin \theta+b \cos \theta=$
  • A
    $\pm \sqrt{a^2+b^2+c^2}$ (b)
  • ✓
    $\pm \sqrt{a^2+b^2-c^2}$
  • C
    $\pm \sqrt{c^2-a^2-b^2}$
  • D
    none of these
Answer
Correct option: B.
$\pm \sqrt{a^2+b^2-c^2}$
B
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MCQ 701 Mark
If $\sin \theta-\cos \theta=0$, then the value of $\sin ^4 \theta+\cos ^4 \theta$ is
  • A
    1
  • B
    $\frac{3}{4}$
  • ✓
    $\frac{1}{2}$
  • D
    $\frac{1}{4}$
Answer
Correct option: C.
$\frac{1}{2}$
C
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MCQ 711 Mark
If $A$ and $B$ are acute angles such that $\sin (A-B)=0$ and $2 \cos (A+B)-1=0$, then $A=$
  • A
    $60^{\circ}$
  • ✓
    $30^{\circ}$
  • C
    $45^{\circ}$
  • D
    $15^{\circ}$
Answer
Correct option: B.
$30^{\circ}$
B
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MCQ 721 Mark
$(\operatorname{cosec} \theta-\sin \theta)(\sec \theta-\cos \theta)(\tan \theta+\cot \theta)$ is equal
  • A
    $0$
  • ✓
    1
  • C
    -1
  • D
    none of these
Answer
Correct option: B.
1
B
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MCQ 731 Mark
The value of $(1+\cot \theta-\operatorname{cosec} \theta)(1+\tan \theta+\sec \theta)$ is
  • A
    1
  • ✓
    2
  • C
    4
  • D
    $0$
Answer
Correct option: B.
2
B
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MCQ 741 Mark
$\cos ^4 A-\sin ^4 A$ is equal to
  • A
    $2 \cos ^2 A+1$
  • ✓
    $2 \cos ^2 A-1$
  • C
    $2 \sin ^2 A-1$
  • D
    $2 \sin ^2 A+1$
Answer
Correct option: B.
$2 \cos ^2 A-1$
B
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MCQ 751 Mark
$\sec ^4 A-\sec ^2 A$ is equal to
  • A
    $\tan ^2 A-\tan ^4 A$
  • B
    $\tan ^4 A-\tan ^2 A$
  • ✓
    $\tan ^4 A+\tan ^2 A$
  • D
    $\tan ^2 A+\tan ^4 A$
Answer
Correct option: C.
$\tan ^4 A+\tan ^2 A$
C
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MCQ 761 Mark
If $\sec \theta+\tan \theta=x$, then $\tan \theta=$
  • A
    $\frac{x^2+1}{x}$
  • B
    $\frac{x^2-1}{x}$
  • C
    $\frac{x^2+1}{2 x}$
  • ✓
    $\frac{x^2-1}{2 x}$
Answer
Correct option: D.
$\frac{x^2-1}{2 x}$
D
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MCQ 771 Mark
If $\sec \theta+\tan \theta=x$, then $\sec \theta=$
  • A
    $\frac{x^2+1}{x}$
  • ✓
    $\frac{x^2+1}{2 x}$
  • C
    $\frac{x^2-1}{2 x}$
  • D
    $\frac{x^2-1}{x}$
Answer
Correct option: B.
$\frac{x^2+1}{2 x}$
B
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MCQ 781 Mark
If $\triangle A B C$ is right angled at $C$, then the value of $\cos (A+B)$ is
  • ✓
    $0$
  • B
    1
  • C
    $\frac{1}{2}$
  • D
    $\frac{\sqrt{3}}{2}$
Answer
Correct option: A.
$0$
A
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MCQ 791 Mark
If $2 \sin ^2 \beta-\cos ^2 \beta=2$, then $\beta$ is equal to
  • A
    $0^{\circ}$
  • ✓
    $90^{\circ}$
  • C
    $45^{\circ}$
  • D
    $30^{\circ}$
Answer
Correct option: B.
$90^{\circ}$
B
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MCQ 801 Mark
If $x=a \sec \theta \cos \phi, y=b \sec \theta \sin \phi$ and $z=c \tan \theta$, then $\frac{x^2}{a^2}+\frac{y^2}{b^2}=$
  • A
    $\frac{z^2}{c^2}$
  • B
    $1-\frac{z^2}{c^2}$
  • C
    $\frac{z^2}{c^2}-1$
  • ✓
    $1+\frac{z^2}{c^2}$
Answer
Correct option: D.
$1+\frac{z^2}{c^2}$
D
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MCQ 811 Mark
If $x=a \sec \theta$ and $y=b \tan \theta$, then $b^2 x^2-a^2 y^2=$
  • A
    $a b$
  • B
    $a^2-b^2$
  • C
    $a^2+b^2$
  • ✓
    $a^2 b^2$
Answer
Correct option: D.
$a^2 b^2$
D
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MCQ 821 Mark
If $x=a \cos \theta$ and $y=b \sin \theta$, then $b^2 x^2+a^2 y^2=$
  • ✓
    $a^2 b^2$
  • B
    $a b$
  • C
    $a^4 b^4$
  • D
    $a^2+b^2$
Answer
Correct option: A.
$a^2 b^2$
A
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MCQ 831 Mark
$\frac{\tan \theta}{\sec \theta-1}+\frac{\tan \theta}{\sec \theta+1}$ is equal to
  • A
    $2 \tan \theta$
  • B
    $2 \sec \theta$
  • ✓
    $2 \operatorname{cosec} \theta$
  • D
    $2 \tan \theta \sec \theta$
Answer
Correct option: C.
$2 \operatorname{cosec} \theta$
C
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MCQ 841 Mark
$\frac{\sin \theta}{1-\cot \theta}+\frac{\cos \theta}{1-\tan \theta}$ is equal to
  • A
    $0$
  • B
    1
  • ✓
    $\sin \theta+\cos \theta$
  • D
    $\sin \theta-\cos \theta$
Answer
Correct option: C.
$\sin \theta+\cos \theta$
C
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MCQ 851 Mark
$\frac{\sin \theta}{1+\cos \theta}$ is equal to
  • A
    $\frac{1+\cos \theta}{\sin \theta}$
  • B
    $\frac{1-\cos \theta}{\cos \theta}$
  • ✓
    $\frac{1-\cos \theta}{\sin \theta}$
  • D
    $\frac{1-\sin \theta}{\cos \theta}$
Answer
Correct option: C.
$\frac{1-\cos \theta}{\sin \theta}$
C
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MCQ 861 Mark
The value of $\sqrt{\frac{1+\cos \theta}{1-\cos \theta}}$ is
  • A
    $\cot \theta-\operatorname{cosec} \theta$
  • ✓
    $\operatorname{cosec} \theta+\cot \theta$
  • C
    $\operatorname{cosec}^2 \theta+\cot ^2 \theta$
  • D
    $(\cot \theta+\operatorname{cosec} \theta)^2$
Answer
Correct option: B.
$\operatorname{cosec} \theta+\cot \theta$
B
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MCQ 871 Mark
$\sqrt{\frac{1+\sin \theta}{1-\sin \theta}}$ is equal to
  • ✓
    $\sec \theta+\tan \theta$
  • B
    $\sec \theta-\tan \theta$
  • C
    $\sec ^2 \theta+\tan ^2 \theta$
  • D
    $\sec ^2 \theta-\tan ^2 \theta$
Answer
Correct option: A.
$\sec \theta+\tan \theta$
A
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MCQ 881 Mark
If $\tan \alpha+\cot \alpha=2$, then $\tan ^{2020} \alpha+\cot ^{2020} \alpha=$
  • A
    $0$
  • ✓
    2
  • C
    2020
  • D
    $2^{2020}$
Answer
Correct option: B.
2
B
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MCQ 891 Mark
If $x=2 \sin ^2 \theta$ and $y=2 \cos ^2 \theta+1$, then $x+y$ is equal to
  • ✓
    3
  • B
    2
  • C
    1
  • D
    $1 / 2$
Answer
Correct option: A.
3
A
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