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
$\cot^2\text{A}$
- ✓
$\tan^2\text{A}$
AnswerCorrect option: D. $\tan^2\text{A}$
$\frac{1+\tan^2\text{A}}{1+\cot^2\text{A}}$
$=\frac{\sec^2\text{A}}{\text{cosec}^2\text{A}}$
$=\frac{\sin^2\text{A}}{\cos^2\text{A}}$
$=\tan^2\text{A}$
View full question & answer→MCQ 21 Mark
$2(\sin^6\theta+\cos^6\theta)-3(\sin^4\theta+\cos^4\theta)$ is equal to:
Answer$2(\sin^6\theta+\cos^6\theta)-3(\sin^4\theta+\cos^4\theta)$
$=2\big[(\sin^2\theta)^3+(\cos^2\theta)^3\big]-3\big[(\sin^2\theta)^2+(\cos^2\theta)^2\big]$
$=2\big[(\sin^2\theta+\cos^2\theta)(\sin^4\theta+\cos^4\theta-\sin^2\theta\cos^2\theta)\big]$
$=-3\big[(\sin^2\theta+\cos^2\theta)^2-2\sin^2\theta\cos^2\theta\big]$
$\{\because\ \text{a}^3+\text{b}^2=(\text{a}+\text{b})^3-3\text{ab}(\text{a}+\text{b})\}$
$=2\big[1(\sin^2\theta)^2+(\cos^2\theta)^2+2\sin^2\theta\cos^2\theta-3\sin^2\theta+\cos^2\theta\big]$
$=-3\big[(1)^2-2\sin^2\theta\cos^2\theta\big]$
$=2\big[(\sin^2\theta+\cos^2\theta)^2-3\sin^2\theta\cos^2\theta\big]-3\big[1-2\sin^2\theta\cos^2\theta\big]$
$=2\big[1-3\sin^2\theta\cos^2\theta\big]-3\big[1-2\sin^2\theta\cos^2\theta\big]$
$=2-6\sin^2\theta\cos^3\theta-3+6\sin^2\theta\cos^2\theta$
$=-1$
View full question & answer→MCQ 31 Mark
If $\text{a}\cos\theta+\text{b}\sin\theta=4\text{ and a}\sin\theta-\text{b}\cos\theta=3,$ then $a^2 + b^2 = 0$
Answer$25$
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).
View full question & answer→MCQ 41 Mark
If $\text{x}=\text{r}\sin\theta\cos\phi,\text{y}=\text{r}\sin\phi\text{ and z}=\text{r}\cos\theta,$ 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
$z^2 + y^2 - x^2 = r^2$
AnswerCorrect 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 $(A).$
View full question & answer→MCQ 51 Mark
$(\sec\text{A}+\tan\text{A})(1-\sin\text{A})=$
- A
$\sec\text{A}$
- B
$\sin\text{A}$
- C
$\text{cosec A}$
- ✓
$\cos\text{A}$
AnswerCorrect 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 $(d).$
View full question & answer→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
$2\tan\theta\sec\theta$
AnswerCorrect 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 $(c).$
View full question & answer→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
$\sin2\alpha$
AnswerCorrect option: B. $\cos2\beta$
$\cos(\alpha+\beta)=0$
$\Rightarrow\ \alpha+\beta=90^\circ \big[\because\ \cos90^\circ=0\big]$
$\Rightarrow\ \alpha=90^\circ-\beta\ .....(\text{i})$
$\sin(\alpha-\beta)=\sin(90^\circ-\beta-\beta)\ \big[\text{using (i)}\big]$
$=\sin\big(90^\circ-2\beta\big)$
$=\cos2\beta\ \big[\because \sin(90^\circ -\theta)=\cos\theta\big]$
View full question & answer→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
$m^2n^2$
- C
$n^2 - m^2$
- ✓
$m^2 + n^2$
AnswerCorrect 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, $a^2 + b^2 = m^2 + n^2.$
View full question & answer→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
$\sin\theta-\cos\theta$
AnswerCorrect option: C. $\sin\theta+\cos\theta$
$\frac{\sin\theta}{1-\cot\theta}+\frac{\cos\theta}{1-\tan\theta}=\frac{\sin\theta}{1-\frac{\cos\theta}{\sin\theta}}+\frac{\cos\theta}{1-\frac{\sin\theta}{\cos\theta}}$
$=\frac{\sin\theta\times\sin\theta}{\sin\theta-\cos\theta}+\frac{\cos\theta\times\cos\theta}{\cos\theta-\sin\theta}$
$=\frac{\sin^2\theta}{\sin\theta-\cos\theta}-\frac{\cos^2\theta}{\sin\theta-\cos\theta}$
$=\frac{\sin^2\theta-\cos^2\theta}{\sin\theta-\cos\theta}$
$=\frac{(\sin\theta+\cos\theta)(\sin\theta-\cos\theta)}{\sin\theta-\cos\theta}$
$=\sin\theta+\cos\theta$
View full question & answer→MCQ 101 Mark
If $\text{x}=\text{a}\sec\theta\cos\phi,\text{y}=\text{b}\sec\theta\sin\phi$ 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
$\frac{\text{z}^2}{\text{c}^2}-1$
- ✓
$1+\frac{\text{z}^2}{\text{c}^2}$
AnswerCorrect 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 $(iv)$
$\frac{\text{x}^2}{\text{a}^2}+\frac{\text{y}^2}{\text{b}^2}-\frac{\text{z}^2}{\text{c}^2}=\sec^2\theta-\tan^2\theta=1$
$\frac{\text{x}^2}{\text{a}^2}+\frac{\text{y}^2}{\text{b}^2}=1+\frac{\text{z}^2}{\text{c}^2}$
View full question & answer→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
$(\cot\theta+\text{cosec }\theta)^2$
AnswerCorrect 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 $(b).$
View full question & answer→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
$\frac{1-\sin\theta}{\cos\theta}$
AnswerCorrect 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 $(C).$
View full question & answer→MCQ 131 Mark
$\frac{\cot\theta}{\cot\theta-\cot3\theta}+\frac{\tan\theta}{\tan\theta-\tan3\theta}$ is equal to:
Answer$=\frac{\cot\theta}{\cot\theta-\cot3\theta}+\frac{\tan\theta}{\tan\theta-\tan3\theta}$
$=\frac{\cot\theta\tan\theta-\cot\theta\tan3\theta+\cot\theta\tan\theta-\tan\theta\cot3\theta}{(\cot\theta-\cot3\theta)(\tan\theta-\tan3\theta)}$
$\{\tan\theta\cot\theta=1\}$
$\Rightarrow\ \frac{1-\cot\theta\tan3\theta+1-\tan\theta\cot3\theta}{\cot\theta\tan\theta-\cot\theta\tan3\theta-\tan\theta\cot3\theta+\cot3\theta\tan3\theta}$
$=\frac{2-\cot\theta\tan3\theta-\tan\theta\cot3\theta}{1-\cot\theta\tan3\theta-\tan\theta\cot3\theta+1}$
$=\frac{2-\cot\theta\tan3\theta-\tan\theta\cot3\theta}{2-\cot\theta\tan\theta-\tan\theta\cot3\theta}$
$=1$
View full question & answer→MCQ 141 Mark
$9\sec^2\text{A}-9\tan^2\text{A}$ is equal to:
AnswerGiven,
$9\sec^2\text{A}-9\tan^2\text{A}$
$=9(\sec^2\text{A}-\tan^2\text{A})$
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 $(b).$
View full question & answer→MCQ 151 Mark
If $\cos9\theta=\sin\theta$ and $9\theta<90^\circ,$ then value of $\tan6\theta$ is:
- A
$\frac{1}{\sqrt{3}}$
- ✓
$\sqrt{3}$
- C
$1$
- D
$0$
AnswerCorrect option: B. $\sqrt{3}$
$\cos(9\theta)=\sin\theta$
$\Rightarrow\ \sin(90^\circ-9\theta)=\sin\theta$
$\Rightarrow\ 90^\circ-9\theta=\theta$
$\Rightarrow\ 90^\circ=\theta+9\theta$
$\Rightarrow\ \theta=10$
$\tan6\theta=\tan6$
$=\tan60^\circ$
$=\sqrt{3}$
View full question & answer→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
$1$
AnswerWe 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 $(b).$
View full question & answer→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
$2\sin^2\text{A}+1$
AnswerCorrect option: B. $2\cos^2\text{A}-1$
$\cos^4\text{A}-\sin^4\text{A}=(\cos^2\text{A}+\sin^2\text{A})(\cos^2\text{A}-\sin^2\text{A})$
$=1(\cos^2\text{A}-\sin^2\text{A})=\cos^2\text{A}-(1-\cos^2\text{A})$
$=\cos^2\text{A}-1+\cos^2\text{A}$
$=2\cos^2\text{A}-1$
View full question & answer→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
$\frac{\text{x}^2-1}{\text{x}}$
AnswerCorrect 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 $(b).$
View full question & answer→MCQ 191 Mark
If $\sin\theta+\sin^2\theta=1$ then $\cos^2\theta+\cos^4\theta$:
Answer$\sin\theta+\sin^2\theta=1$
$\Rightarrow\ \sin\theta=1-\sin^2\theta$
$\Rightarrow\ \sin\theta=\cos^2\theta$
$\cos^2\theta+\cos^4\theta=\sin\theta+\sin^2\theta\ \{\because \cos^2\theta=\sin\theta\}$
$\Rightarrow\ \cos^2\theta+\cos^4\theta=1$
$\{\because \sin\theta+\sin^2\theta=1(\text{given})\}$
View full question & answer→MCQ 201 Mark
The value of $(1+\cot\theta-\text{cosec }\theta)(1+\tan\theta+\sec\theta)$ is
Answer$(1+\cot\theta-\text{cosec }\theta)(1+\tan\theta+\sec\theta)$
$=\Big(1+\frac{\cos\theta}{\sin\theta}-\frac{1}{\sin\theta}\Big)\Big(1+\frac{\sin\theta}{\cos\theta}+\frac{1}{\cos\theta}\Big)$
$=\frac{(\sin\theta+\cos\theta-1)(\cos\theta+\sin\theta+1)}{\sin\theta\times\cos\theta}$
$=\frac{\{(\sin\theta+\cos\theta)-1\}\{(\cos\theta+\sin\theta)+1\}}{\sin\theta\cos\theta}$
$=\frac{(\cos\theta+\sin\theta)^2-1}{\sin\theta\cos\theta}$
$=\frac{\cos^2\theta+\sin^2\theta+2\sin\theta\cos\theta-1}{\sin\theta\cos\theta}$
$=\frac{1+2\sin\theta\cos\theta-1}{\sin\theta\cos\theta}$
$=\frac{2\sin\theta\cos\theta}{\sin\theta\cos\theta}$
$=2$
View full question & answer→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
$b - a$
AnswerCorrect 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,
$\text{p}^2-\text{q}^2=(\text{a}\cot\theta+\text{b cosec }\theta)^2-(\text{b}\cot\theta+\text{a cosec }\theta)^2$
$=\text{a}^2\cot^2\theta+\text{b}^2\text{cosec}^2\theta+2\text{ab}\cot\theta\text{ cosec }\theta$
$=-(\text{b}^2\cot^2\theta+\text{a}^2\text{cosec}^2\theta+2\text{ab}\cot\theta\text{ cosec }\theta)$
$=\text{a}^2\cot^2\theta+\text{b}^2\text{cosec}^2\theta+2\text{ab}\cot\theta\text{ cosec }\theta$
$=-\text{b}^2\cot^2\theta-\text{a}^2\text{cosec}^2\theta-2\text{ab}\cot\theta\text{ cosec }\theta$
$=\text{a}^2(\cot^2\theta-\text{cosec}^2\theta)+\text{b}^2(\text{cosec}^2\theta-\cot^2\theta)$
$=-\text{a}^2(\text{cosec}^2\theta-\cot^2\theta)+\text{b}^2(\text{cosec}^2\theta-\cot^2\theta)$
$=-\text{a}^2\times1+\text{b}^2\times1$
$=\text{b}^2-\text{a}^2$
View full question & answer→MCQ 221 Mark
$(\text{cosec }\theta-\sin\theta)(\sec\theta-\cos\theta)(\tan\theta+\cot\theta)$ is equal to:
Answer$(\text{cosec }\theta-\sin\theta)(\sec\theta-\cos\theta)(\tan\theta+\cot\theta)$
$=\Big(\frac{1}{\sin\theta}-\sin\theta\Big)\Big(\frac{1}{\cos\theta}-\cos\theta\Big)$
$\Big(\frac{\sin\theta}{\cos\theta}+\frac{\cos\theta}{\sin\theta}\Big)$
$=\frac{1-\sin^2\theta}{\sin\theta}\times\frac{1-\cos^2\theta}{\cos\theta}\times\frac{\sin^2\theta+\cos^2\theta}{\sin\theta\cos\theta}$
$=\frac{\cos^2\theta}{\sin\theta}\times\frac{\sin^2\theta}{\cos\theta}\times\frac{1}{\sin\theta\cos\theta}$
$=\frac{\sin^2\theta\cos^2\theta}{\sin^2\theta\cos^2\theta}$
$=1$
View full question & answer→MCQ 231 Mark
$\sqrt{\frac{1+\sin\theta}{1-\sin\theta}}$ is equal to:
AnswerCorrect option: A. $\sec\theta+\tan\theta$
$\sqrt{\frac{1+\sin\theta}{1-\sin\theta}}=\sqrt{\frac{(1+\sin\theta)(1+\sin\theta)}{(1-\sin\theta)(1+\sin\theta)}}$
$=\sqrt{\frac{(1+\sin\theta)^2}{1-\sin^2\theta}}$
$=\sqrt{\frac{(1+\sin\theta)^2}{\cos^2\theta}}$
$=\frac{1+\sin\theta}{\cos\theta}$
$=\frac{1}{\cos\theta}+\frac{\sin\theta}{\cos\theta}$
$=\sec\theta+\tan\theta$
View full question & answer→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
$\tan^2\text{A}+\tan^4\text{A}$
AnswerCorrect option: C. $\tan^4\text{A}+\tan^2\text{A}$
$\sec^4-\sec^2\text{A}=\sec^2\text{A}(\sec^2\text{A}-1)$
$=(1+\tan^2\text{A})\tan^2\text{A}$
$\begin{cases}\sec^2\text{A}=1+\tan^2\text{A}\\
\sec^2\text{A}-1=\tan^2\text{A}\end{cases}$
$=\tan^2\text{A}+\tan^4\text{A}$
$=\tan^4\text{A}+\tan^2\text{A}$
View full question & answer→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
$\frac{\text{x}^2+1}{2\text{x}}$
- ✓
$\frac{\text{x}^2-1}{2\text{x}}$
AnswerCorrect 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 $(i)$
$2\tan\theta=\text{x}-\frac{1}{\text{x}}=\frac{\text{x}^2-1}{\text{x}}$
$\tan\theta=\frac{\text{x}^2-1}{2\text{x}}$
View full question & answer→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
$\frac{\sqrt{3}}{2}$
AnswerIn a right angled traingle $\text{ABC}, \triangle\text{C}$ is a righta angle.
We know that, the sum of angles of a triangle is $180^\circ$.
$\therefore\ \angle\text{A}+\angle\text{B}+\angle\text{C}=180^\circ$
$\Rightarrow\ \angle\text{A}+\angle\text{B}+90^\circ=180^\circ$
$\Rightarrow\ \angle\text{A}+\angle\text{B}=90^\circ$
$\therefore\ \cos(\text{A}+\text{B})=\cos90^\circ=0$
Hence, the correct answer is option $(a).$
View full question & answer→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
AnswerCorrect option: B. $\pm\sqrt{\text{a}^2+\text{b}^2-\text{c}^2}$
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 $
$\ (\text{Dividing by}-1)$
$(\text{a}\sin\theta+\text{b}\cos\theta)^2=\text{a}^2+\text{b}^2-\text{c}^2$
$\therefore\ \text{a}\sin\theta+\text{b}\cos\theta=\pm\sqrt{\text{a}^2+\text{b}^2-\text{c}^2}$
View full question & answer→MCQ 281 Mark
If $\text{x}=\text{a}\cos\theta\text{ and y}=\text{b}\sin\theta,$ then $b^2x^2 + a^2y^2 =$
- ✓
$a^2b^2$
- B
$ab$
- C
$a^4b^4$
- D
$a^2 + b^{_2}$
AnswerCorrect option: A. $a^2b^2$
$a^2b^2$
$\text{x}=\text{a}\cos\theta,\text{y}=\text{b}\sin\theta\ .....(\text{i})$
$\text{bx}=\text{ab}\cos\theta,\text{ay}=\text{ab}\sin\theta\ .....(\text{ii})$
Adding (i) and (ii) we get,
$=\text{b}^2\text{x}^2+\text{a}^2\text{b}^2\cos^2\theta+\text{a}^2\text{b}^2\sin^2\theta$
$=\text{a}^2\text{b}^2(\cos^2\theta+\sin^2\theta)$
$=\text{a}^2\text{b}^2\times1$
$=\text{a}^2\text{b}^2$
View full question & answer→MCQ 291 Mark
If $\text{x}=\text{a}\sec\theta \text{ and y}=\text{b}\tan\theta,$ then $b^2x^2 - a^2y^2 $=
- A
$ab$
- B
$a^2 - b^2$
- C
$a^2 + b^2$
- ✓
$a^2b^2$
AnswerCorrect option: D. $a^2b^2$
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 $(D).$
View full question & answer→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$\sin^2{29}^\circ+\sin^2{61}^\circ$
$=\sin^2{29}^\circ+\sin^2{(90^\circ-29^\circ})$
$=\sin^2{29}+\cos^2{29}^\circ$
$(\sin^2\theta+\cos^2\theta=1)$
View full question & answer→MCQ 311 Mark
$(1+\tan\theta+\sec\theta)(1+\cot\theta-\text{cosec }\theta)=$
Answer$(1+\tan\theta+\sec\theta)(1+\cot\theta-\text{cosec }\theta)$
$=1+\cot\theta-\text{cosec }\theta+\tan\theta+\cot\theta\tan\theta$
$=-\tan\theta\text{ cosec }\theta+\sec\theta+\sec\theta\cot\theta-\sec\theta\text{ cosec }\theta$
$=1+\frac{\cos\theta}{\sin\theta}-\frac{1}{\sin\theta}+\frac{\sin\theta}{\cos\theta}+1-\frac{\sin\theta}{\cos\theta}\times\frac{1}{\sin\theta}$
$=\frac{1}{\cos\theta}+\frac{1}{\cos\theta}\times\frac{\cos\theta}{\sin\theta}-\frac{1}{\cos\theta}\times\frac{1}{\sin\theta}$
$=2+\frac{\cos\theta}{\sin\theta}+\frac{\sin\theta}{\cos\theta}-\frac{1}{\sin\theta}-\frac{1}{\cos\theta}$
$=\frac{1}{\cos\theta}+\frac{1}{\sin\theta}-\frac{1}{\sin\theta\cos\theta}$
$=2+\frac{\sin\theta}{\cos\theta}+\frac{\cos\theta}{\sin\theta}-\frac{1}{\sin\theta\cos\theta}$
$=2+\frac{\sin^2\theta+\cos^2\theta}{\sin\theta\cos\theta}-\frac{1}{\sin\theta\cos\theta}$
$=2+\frac{1}{\sin\theta\cos\theta}-\frac{1}{\sin\theta\cos\theta}$
$=2$
View full question & answer→MCQ 321 Mark
If $\cos\text{A}+\cos^2\text{A}=1,$ then $\sin^2\text{A}+\sin^4\text{A}=$
AnswerGiven,
$\cos\text{A}+\cos^2\text{A}=1$
$\Rightarrow\ 1-\cos^2\text{A}=\cos\text{A}$
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 $(c).$
View full question & answer→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
$\frac{1}{4}$
AnswerCorrect 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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