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Maths M.C.Q (1 Marks)

Trigonometric Identities · Maharashtra Board STD 10 (MSBSHSE - English Medium). 57 questions.

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M.C.Q (1 Marks)

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57 questions · auto-graded multiple-choice test.

MCQ 11 Mark
$\tan10^\circ\tan15^\circ\tan75^\circ\tan80^\circ=?$
  • A
    $\sqrt{3}$
  • B
    $\frac{1}{\sqrt{3}}$
  • C
    $-1$
  • ✓
    $1$
Answer
Correct option: D.
$1$
$\tan10^\circ\tan15^\circ\tan75^\circ\tan80^\circ$
$=\tan10^\circ\tan15^\circ\tan(90^\circ-15^\circ)\tan(90^\circ-10^\circ)$
$=\tan10^\circ\tan15^\circ\cot15^\circ\cot10^\circ$
$=(\tan10^\circ\cot10^\circ)(\tan15^\circ\cot15^\circ)$
$=1\times1$
$=1$
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MCQ 21 Mark
If $\sec\theta=\frac{25}{7}$ then $\sin\theta=?$
  • A
    $\frac{7}{24}$
  • B
    $\frac{24}{7}$
  • ✓
    $\frac{24}{25}$
  • D
    None of these.
Answer
Correct option: C.
$\frac{24}{25}$
Consider $\triangle\text{ABC}$ where $\angle\text{B}=90^\circ,\angle\text{A}=\theta.$
Then, $\sec\theta=\frac{\text{Hypotenuse}}{\text{Base}}$
$=\frac{\text{AC}}{\text{AB}}=\frac{25}{7}$
Let $\text{AC}=25\text{k}$ and $\text{AB}=7\text{k},$
where $k$ is positive.
By Pythagoras Theorem,
$\text{AC}^2=\text{AB}^2+\text{BC}^2$
$\Rightarrow(25\text{k})^2=(7\text{k})^2+\text{BC}^2$
$\Rightarrow\text{BC}^2=\text{625k}^2-\text{49k}^2=\text{576k}^2$
$\Rightarrow\text{BC}=\text{24k}$
Now, $\sin\theta=\frac{\text{Perpendicular}}{\text{Hypotenuse}}$
$\frac{\text{BC}}{\text{AC}}=\frac{24\text{k}}{\text{25k}}=\frac{24}{25}$
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MCQ 31 Mark
If $3\cot\theta=4$ then $\frac{(5\sin\theta+3\cos\theta)}{(5\sin\theta-3\cos\theta)} =?$
  • A
    $\frac{1}{3}$
  • B
    $3$
  • C
    $\frac{1}{9}$
  • ✓
    $9$
Answer
Correct option: D.
$9$
Given, $3\cot\theta=4$
Now, $\frac{(5\sin\theta+3\cos\theta)}{(5\sin\theta-3\cos\theta)} $
$=\frac{\frac{5\sin\theta}{\sin\theta}+\frac{3\cos\theta}{\sin\theta}}{\frac{5\sin\theta}{\sin\theta}-\frac{3\cos\theta}{\sin\theta}}$
$=\frac{5+3\cot\theta}{5-3\cot\theta}$
$=\frac{5+4}{5-4}$
$=\frac{9}{1}$
$=9$
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MCQ 41 Mark
If $7\tan\theta=4$ then $\frac{(7\sin\theta-3\cos\theta)}{(7\sin\theta+3\cos\theta)} =?$
  • ✓
    $\frac{1}{7}$
  • B
    $\frac{5}{7}$
  • C
    $\frac{3}{7}$
  • D
    $\frac{5}{14}$
Answer
Correct option: A.
$\frac{1}{7}$
Given, $7\tan\theta=4$
Now, $\frac{(7\sin\theta-3\cos\theta)}{(7\sin\theta+3\cos\theta)} $
$=\frac{\frac{7\sin\theta}{\cos\theta}-\frac{3\cos\theta}{\cos\theta}}{\frac{7\sin\theta}{\cos\theta}+\frac{3\cos\theta}{\cos\theta}}$
$=\frac{7\tan\theta-3}{7\tan\theta+3}$
$=\frac{4-3}{4+3}$
$=\frac{1}{7}$
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MCQ 51 Mark
If $\cos\theta=\frac{4}{5}$ then $\tan\theta=?$
  • ✓
    $\frac{3}{4}$
  • B
    $\frac{4}{3}$
  • C
    $\frac{3}{5}$
  • D
    $\frac{5}{3}$
Answer
Correct option: A.
$\frac{3}{4}$
$\sin^2\theta+\cos^2\theta=1$
$\Rightarrow\sin^2\theta+\Big(\frac{4}5{}\Big)^2=1$
$\Rightarrow\sin^2\theta=1-\frac{16}{25}=\frac{9}{25}$
$\Rightarrow\sin\theta=\frac{3}{5}$
$\therefore\tan\theta=\frac{\sin\theta}{\cos\theta}$
$=\frac{\frac{3}{5}}{\frac{4}{5}}=\frac{3}{4}$
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MCQ 61 Mark
$\sec^210^\circ-\cot^280^\circ=?$
  • ✓
    $1$
  • B
    $0$
  • C
    $\frac{3}{2}$
  • D
    $\frac{1}2{}$
Answer
Correct option: A.
$1$
$\sec^210^\circ-\cot^280^\circ$
$=\sec^210^\circ-\cot^2(90^\circ-10^\circ)$
$=\sec^210^\circ-\tan^210^\circ$
$=1$
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MCQ 71 Mark
If $\sec\text{4A}=\text{cosec}(\text{A}-10^\circ)$ and $\text{4A}$ is acute then $\angle\text{A}=?$
  • ✓
    $20^\circ$
  • B
    $30^\circ$
  • C
    $40^\circ$
  • D
    $50^\circ$
Answer
Correct option: A.
$20^\circ$
$\sec\text{4A}=\text{cosec}(\text{A}-10^\circ)$
$\Rightarrow\text{cosec}(90^\circ-\text{4A})=\text{cosec}(\text{A}-10^\circ)$
$\Rightarrow90^\circ-\text{4A}=\text{A}-10^\circ$
$\Rightarrow\text{5A}=100^\circ$
$\Rightarrow\text{A}=20^\circ$
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MCQ 81 Mark
The value of $\Big(\sin^230^\circ\cos^245^\circ+4\tan^230^\circ+\frac{1}{2}\sin^290^\circ+\frac{1}{8}\cot^260^\circ\Big)=?$
  • A
    $\frac{3}8{}$
  • B
    $\frac{5}{8}$
  • C
    $6$
  • ✓
    $2$
Answer
Correct option: D.
$2$
$\Big(\sin^230^\circ\cos^245^\circ+4\tan^230^\circ+\frac{1}{2}\sin^290^\circ+\frac{1}{8}\cot^260^\circ\Big)$
$=\frac{1}{2^2}\times\frac{1}{\big(\sqrt2\big)^2}+4\times\frac{1}{\big(\sqrt3\big)^2}+\frac{1}{2}+1^2+\frac{1}{8}\times\frac{1}{\big(\sqrt3\big)^2}$
$\begin{bmatrix}\because\sin30^\circ=\frac{1}{2}\ \text{and }\cos45^\circ=\frac{1}{\sqrt2}\\\text{and }\tan30^\circ=\frac{1}{2 }\ \text{and }\cot60^\circ=\frac{1}{\sqrt3}\end{bmatrix}$
$=\frac{1}{4}\times\frac{1}{2}+4\times\frac{1}{3}+\frac{1}{2}+\frac{1}{24}$
$=\frac{1}{8}+\frac{4}{3}+\frac{1}{2}+\frac{1}{24}$
$=\frac{3+32+12+1}{24}$
$=\frac{48}{24}$
$=2$
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MCQ 91 Mark
If $\sin\text{3A}=\cos(\text{A}-10^\circ)$ and $\text{3A}$ is acute then $\angle\text{A}=?$
  • A
    $35^\circ$
  • ✓
    $25^\circ$
  • C
    $20^\circ$
  • D
    $45^\circ$
Answer
Correct option: B.
$25^\circ$
$\sin\text{3A}=\cos(\text{A}-10^\circ)$
$\Rightarrow\cos(90^\circ-\text{3A})=\cos(\text{A}-10^\circ)$
$\Rightarrow90^\circ-\text{3A}=\text{A}-10^\circ$
$\Rightarrow\text{4A}=100^\circ$
$\Rightarrow\text{A}=25^\circ$
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MCQ 101 Mark
$\sec70^\circ\sin20^\circ+\cos20^\circ\text{cosec}\ 70^\circ=?$
  • A
    $0$
  • B
    $1$
  • C
    $-1$
  • ✓
    $2$
Answer
Correct option: D.
$2$
$\sec70^\circ\sin20^\circ+\cos20^\circ\text{cosec}70^\circ$
$=\frac{1}{\cos70^\circ}\times\sin20^\circ+\cos20^\circ\times\frac{1}{\sin70^\circ}$
$=\frac{1}{\cos(90^\circ-20^\circ)}\times\sin20^\circ+\cos20^\circ\times\frac{1}{\sin(90^\circ-20^\circ)}$
$=\frac{1}{\sin20^\circ}\times\sin20^\circ+\cos20^\circ\times\frac{1}{\cos20^\circ}$
$=1+1$
$=2$
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MCQ 111 Mark
If $\cos(\alpha+\beta)=0$ then $\sin(\alpha-\beta)=?$
  • A
    $\sin\alpha$
  • B
    $\cos\beta$
  • C
    $\sin2\alpha$
  • ✓
    $\cos2\beta$
Answer
Correct option: D.
$\cos2\beta$
$\cos(\alpha+\beta)=0$
$\Rightarrow\cos(\alpha+\beta)=\cos90^\circ$
$\Rightarrow\alpha+\beta=90^\circ$
$\Rightarrow\alpha=90^\circ-\beta$
Now, $\sin(\alpha-\beta)=\sin(90^\circ-\beta-\beta)$
$=\sin(90^\circ-2\beta)$
$=\cos2\beta$
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MCQ 121 Mark
If $\sin\theta=\frac{\text{a}}{\text{b}}$ then $\cos\theta=?$
  • A
    $\frac{\text{a}}{\sqrt{\text{b}^2-\text{a}^2}}$
  • ✓
    $\frac{\sqrt{\text{b}^2-\text{a}^2}}{\text{b}}$
  • C
    $\frac{\text{a}}{\sqrt{\text{b}^2-\text{a}^2}}$
  • D
    $\frac{\text{b}}{\text{a}}$
Answer
Correct option: B.
$\frac{\sqrt{\text{b}^2-\text{a}^2}}{\text{b}}$
$\cos^2\theta=1-\sin^2\theta$
$=1-\Big(\frac{\text{a}}{\text{b}}\Big)^2$
$=1-\frac{\text{a}^2}{\text{b}^2}$
$=\frac{\text{b}^2-\text{a}^2}{\text{b}^2}$
$\Rightarrow\cos\theta=\frac{\sqrt{\text{b}^2-\text{a}^2}}{\text{b}}$
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MCQ 131 Mark
$\Bigg\{\frac{\big(\sin^222^\circ+\sin^268^\circ\big)}{\big(\cos^222^\circ+\cos^268^\circ\big)}+\sin^263^\circ
+\cos63^\circ\sin27^\circ\Bigg\}=?$
  • A
    $0$
  • B
    $1$
  • ✓
    $2$
  • D
    $3$.
Answer
Correct option: C.
$2$
$=\Bigg\{\frac{\big(\sin^222^\circ+\sin^268^\circ\big)}{\big(\cos^222^\circ+\cos^268^\circ\big)}+\sin^263^\circ +\cos63^\circ\sin27^\circ\Bigg\}$
$=\Bigg\{\frac{\big(\sin^222^\circ+\sin^2(90^\circ-22^\circ)\big)}{\big(\cos^222^\circ+\cos^2(90^\circ-22^\circ)\big)}+\sin^263^\circ+\cos63^\circ\sin(90^\circ-63^\circ)\Bigg\}$
$=\Bigg\{\frac{\big(\sin^222^\circ+\cos^222^\circ\big)}{\big(\cos^222^\circ+\sin^222^\circ\big)}+\sin^263^\circ+\cos63^\circ\cos63^\circ\Bigg\}$
$=\Bigg\{\frac{1}{1}+\sin^263^\circ+\cos^263^\circ\Bigg\}$
$=1+1$
$=2$
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MCQ 141 Mark
If $\sin\theta=\frac{\sqrt3}{2}$ then $(\text{cosec }\theta+\cot\theta)=?$
  • A
    $\big(2+\sqrt3\big)$
  • B
    $2\sqrt3$
  • C
    $\sqrt2$
  • ✓
    $\sqrt3$
Answer
Correct option: D.
$\sqrt3$
Given $\sin\theta=\frac{\sqrt3}{2}$ and $\text{cosec }\theta=\frac{2}{\sqrt3}$
$\text{cosec}^2\theta-\cot^2\theta=1$
$\Rightarrow\cot^2\theta=\text{cosec}^2\theta-1$
$\Rightarrow\cot^2\theta=\frac{4}3{}-1 [$Given$]$
$\Rightarrow\cot\theta=\frac{1}{\sqrt3}$
$\therefore\text{cosec }\theta+\cot\theta=\frac{2}{\sqrt3}+\frac{1}{\sqrt3}$
$=\frac{2}{\sqrt3}$
$=\frac{\sqrt3\times\sqrt3}{\sqrt3}$
$=\sqrt3$
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MCQ 151 Mark
$\frac{\cot(90^\circ-\theta).\sin(90^\circ-\theta)}{\sin\theta}+\frac{\cot40^\circ}{\tan50^\circ}-\big(\cos^220^\circ+\cos^270^\circ\big)=? $
  • A
    $0$
  • ✓
    $1$
  • C
    $-1$
  • D
    None of these.
Answer
Correct option: B.
$1$
$\frac{\cot(90^\circ-\theta).\sin(90^\circ-\theta)}{\sin\theta}+\frac{\cot40^\circ}{\tan50^\circ}-\big(\cos^220^\circ+\cos^270^\circ\big)$
$=\frac{\tan\theta\times\cos\theta}{\sin\theta}+\frac{\cot40^\circ}{\tan(90^\circ-40^\circ)}-\big(\cos^220^\circ+\cos^2(90^\circ-20^\circ)\big)$
$=\frac{\frac{\sin\theta}{\cos\theta}\times\cos\theta}{\sin\theta}+\frac{\cot40^\circ}{\cot40^\circ}-\big(\cos^220^\circ+\sin^220^\circ\big)$
$=\frac{\sin\theta}{\sin\theta}+1-1$
$=1+0$
$=1$
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MCQ 161 Mark
If $\text{x}\tan45^\circ\cos60^\circ=\sin60^\circ\cot60^\circ$ then $\text{x}=?$
  • ✓
    $1$
  • B
    $\frac{1}{2}$
  • C
    $\frac{1}{\sqrt2}$
  • D
    $\sqrt3$
Answer
Correct option: A.
$1$
$\text{x}\tan45^\circ\cos60^\circ=\sin60^\circ\cot60^\circ$
$\Rightarrow\text{x}\times1\times\frac{1}{2}=\frac{\sqrt3}{2}\times\frac{1}{\sqrt3}$
$\Rightarrow\frac{\text{x}}2{}=\frac{1}{2}$
$\Rightarrow\text{x}=1$
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MCQ 171 Mark
$3\cos^260^\circ+2\cot^230^\circ-5\sin^245^\circ=?$
  • A
    $\frac{13}{6}$
  • ✓
    $\frac{17}{4}$
  • C
    $1$
  • D
    $4$
Answer
Correct option: B.
$\frac{17}{4}$
$3\cos^260^\circ+2\cot^230^\circ-5\sin^245^\circ$
$=3\times\Big(\frac{1}{2}\Big)^2+2\big(\sqrt3\big)^2-5\Big(\frac{1}{\sqrt2}\Big)^2$
$=3\times\frac{1}{4}+2\times3-5\times\frac{1}{2}$
$=\frac{3}{4}+6-\frac{5}{2}$
$=\frac{3+24-10}{4}$
$=\frac{17}{4}$
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MCQ 181 Mark
If $\tan\theta=\frac{\text{a}}{\text{b}}$ then $\frac{(\cos\theta+\sin\theta)}{(\cos\theta-\sin\theta)}=?$
  • A
    $\frac{\text{a}+\text{b}}{\text{a}-\text{b}}$
  • B
    $\frac{\text{a}-\text{b}}{\text{a}+\text{b}}$
  • ✓
    $\frac{\text{b}+\text{a}}{\text{b}-\text{a}}$
  • D
    $\frac{\text{b}-\text{a}}{\text{b}+\text{a}}$
Answer
Correct option: C.
$\frac{\text{b}+\text{a}}{\text{b}-\text{a}}$
Given, $\tan\theta=\frac{\text{a}}{\text{b}}$
Now, $\frac{(\cos\theta+\sin\theta)}{(\cos\theta-\sin\theta)}=\frac{\frac{\cos\theta}{\cos\theta}+\frac{sin\theta}{\cos\theta}}{\frac{\cos\theta}{\cos\theta}-\frac{\sin\theta}{\cos\theta}}$
$=\frac{1+\tan\theta}{1-\tan\theta}$
$=\frac{1+\frac{\text{a}}{\text{b}}}{1-\frac{\text{a}}{\text{b}}}$
$=\frac{\frac{\text{b}+\text{a}}{\text{b}}}{\frac{\text{b}-\text{a}}{\text{b}}}$
$=\frac{\text{b}+\text{a}}{\text{b}-\text{a}}$
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MCQ 191 Mark
If $2\cos3\theta=1$ then $\theta=?$
  • A
    $10^\circ$
  • B
    $15^\circ$
  • ✓
    $20^\circ$
  • D
    $30^\circ$
Answer
Correct option: C.
$20^\circ$
$2\cos3\theta=1$
$\Rightarrow\cos3\theta=\frac{1}{2}$
$\Rightarrow\cos3\theta=\cos60^\circ$
$\Rightarrow3\theta=\cos60^\circ$
$\Rightarrow\theta=20^\circ$
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MCQ 201 Mark
$\frac{\tan35^\circ}{\cot55^\circ}+\frac{\cot78^\circ}{\tan12^\circ}=?$
  • A
    $0$
  • B
    $1$
  • ✓
    $2$
  • D
    None of these.
Answer
Correct option: C.
$2$
$\frac{\tan35^\circ}{\cot55^\circ}+\frac{\cot78^\circ}{\tan12^\circ}$
$=\frac{\tan(90^\circ-55^\circ)}{\cot55^\circ}+\frac{\cot78^\circ}{\tan(90^\circ-78^\circ)}$
$=\frac{\cot55^\circ}{\cot55^\circ}+\frac{\cot78^\circ}{\cot78^\circ}$
$=1+1$
$=2$
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MCQ 211 Mark
If $\tan\theta=\sqrt3$ then $\sec\theta=?$
  • A
    $\frac{2}{\sqrt3}$
  • B
    $\frac{\sqrt3}{2}$
  • C
    $\frac{1}{2}$
  • ✓
    $2$
Answer
Correct option: D.
$2$
Consider $\triangle\text{ABC}$ where $\angle\text{B}=90^\circ,\angle\text{A}=\theta.$
Then, $\tan\theta=\frac{\text{perpendicular}}{\text{Base}}$
$=\frac{\text{BC}}{\text{AB}}=\frac{\sqrt3}{1}$
Let $\text{BC}=\sqrt{3}\text{k}$ and $\text{AB}=\text{k},$
where $k$ is positive.
By Pythagoras Theorem,
$\text{AC}^2=\text{AB}^2+\text{BC}^2$
$\Rightarrow\text{AC}^2=(\text{k})^2+(\sqrt3\text{k})^2=\text{k}^2+3\text{k}^2=4\text{k}^2$
$\Rightarrow\text{AC}=\text{2k}$
Now, $\sec\theta=\frac{\text{Hypotenuse}}{\text{Base}}$
$\frac{\text{AC}}{\text{AB}}=\frac{2\text{k}}{\text{k}}=2$
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MCQ 221 Mark
If $\sqrt3\tan2\theta-3=0$ then $\theta=?$
  • A
    $15^\circ$
  • ✓
    $30^\circ$
  • C
    $45^\circ$
  • D
    $60^\circ$
Answer
Correct option: B.
$30^\circ$
$\sqrt3\tan2\theta-3=0$
$\Rightarrow\sqrt3\tan2\theta-3$
$\Rightarrow\tan2\theta=\frac{3}{\sqrt3}$
$\Rightarrow\tan2\theta=\sqrt3$
$\Rightarrow\tan2\theta=\tan60^\circ$
$\Rightarrow2\theta=60^\circ$
$\Rightarrow\theta=30^\circ$
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MCQ 241 Mark
$\frac{\cos38^\circ\text{cosec}52^\circ}{\tan18^\circ\tan35^\circ\tan60^\circ\tan72^\circ\tan55^\circ}=?$
  • A
    $\sqrt3$
  • B
    $\frac{1}{3}$
  • ✓
    $\frac{1}{\sqrt3}$
  • D
    $\frac{2}{\sqrt3}$
Answer
Correct option: C.
$\frac{1}{\sqrt3}$
$\frac{\cos38^\circ\text{cosec}52^\circ}{\tan18^\circ\tan35^\circ\tan60^\circ\tan72^\circ\tan55^\circ}$
$=\frac{\cos(90^\circ-52^\circ)\text{cosec}52^\circ}{\tan18^\circ\times\tan35^\circ\times\sqrt3\times\tan(90^\circ-18^\circ)\times\tan(90^\circ-35^\circ)}$
$=\frac{\sin52^\circ\text{cosec}52^\circ}{\tan18^\circ\times\tan35^\circ\times\sqrt3\times\sec18^\circ\times\sec35^\circ}$
$=\frac{\sin52^\circ\times\frac{1}{\sin52^\circ}}{(\tan18^\circ\sec18^\circ)\times\sqrt3\times(\tan35^\circ\sec35^\circ)}$
$=\frac{1}{1\times\sqrt3\times1}$
$=\frac{1}{\sqrt3}$
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MCQ 251 Mark
If $(\cos\theta+\sec\theta)=\frac{5}{2}$ then $(\cos^2\theta+\sec^2\theta)=?$
  • A
    $\frac{21}{4}$
  • ✓
    $\frac{17}{4}$
  • C
    $\frac{29}{4}$
  • D
    $\frac{33}{4}$
Answer
Correct option: B.
$\frac{17}{4}$
$(\cos\theta+\sec\theta)=\frac{5}{2}$
$\Rightarrow(\cos\theta+\sec\theta)^2=\Big(\frac{5}{2}\Big)^2$
$\Rightarrow\cos^2\theta+\sec^2\theta+2\cos\theta\sec\theta=\frac{25}{4}$
$\Rightarrow\cos^2\theta+\sec^2\theta+2\cos\theta\times\frac{1}{\cos\theta}=\frac{25}{4}$
$\Rightarrow\cos^2\theta+\sec^2\theta+2=\frac{25}{4}$
$\Rightarrow\cos^2\theta+\sec^2\theta=\frac{25}{4}-2$
$\Rightarrow\cos^2\theta+\sec^2\theta=\frac{25-8}{4}$
$\Rightarrow\cos^2\theta+\sec^2\theta=\frac{17}{4}$
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MCQ 261 Mark
$\frac{\cos^256^\circ+\cos^234^\circ}{\sin^256^\circ+\sin^234^\circ}+3\tan^256^\circ\tan^234^\circ=?$
  • A
    $3\frac{1}2{}$
  • ✓
    $4$
  • C
    $6$
  • D
    $5$
Answer
Correct option: B.
$4$
$\frac{\cos^256^\circ+\cos^234^\circ}{\sin^256^\circ+\sin^234^\circ}+3\tan^256^\circ\tan^234^\circ$
$=\frac{\big\{\cos(90^\circ-34^\circ)\big\}^2+\cos^234^\circ}{\big\{\sin(90^\circ-34^\circ)\big\}^2+\sin^234^\circ}+3\big\{\tan(90^\circ-34^\circ)\big\}^2\tan^234^\circ$
$=\frac{\sin^234^\circ+\cos^234^\circ}{\cos^234^\circ+\sin^234^\circ}+3\cot^234^\circ\tan^234^\circ$ $\begin{bmatrix}\because\cos(90^\circ-\theta)=\sin\theta,\sin(90^\circ-\theta)\\=\cos\theta\ \text{and }\tan(90^\circ-\theta)=\cot\theta\end{bmatrix}$
$=\frac{1}{1}+3\times1$
$\Big[\because\cot\theta=\frac{1}{\tan\theta}\ \text{and }\sin^2\theta+\cos^2\theta=1\Big]$
$=4$
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MCQ 271 Mark
If $\text{cosec}\theta=\sqrt{10}$ then $\sec\theta=?$
  • A
    $\frac{3}{\sqrt{10}}$
  • ✓
    $\frac{\sqrt{10}}{3}$
  • C
    $\frac{1}{\sqrt{10}}$
  • D
    $\frac{2}{\sqrt{10}}$
Answer
Correct option: B.
$\frac{\sqrt{10}}{3}$
Consider $\triangle\text{ABC}$
where $\angle\text{B}=90^\circ,\angle\text{A}=\theta.$
Then, $\text{cosec}\theta=\frac{\text{Hypotenuse}}{\text{perpendicular}}$
$=\frac{\text{AC}}{\text{BC}}=\frac{\sqrt{10}}{1}$
Let $\text{AC}=\sqrt{10}\text{k}$ and $\text{BC}=\text{k}$
where $k$ is positive.
By Pythagoras Theorem,
$\text{AC}^2=\text{AB}^2+\text{BC}^2$
$\Rightarrow\big(\sqrt{10}\text{k}\big)^2=\text{AB}^2+(\text{k})^2$
$\Rightarrow\text{AB}^2=10\text{k}^2-\text{k}^2=\text{9k}^2$
$\Rightarrow\text{AB}$
$=\text{3k}$
$\sec\theta=\frac{\text{Hypotenuse}}{\text{Base}}=\frac{\text{AC}}{\text{AB}}$
$=\frac{\sqrt{10}\text{k}}{\text{3k}}$
$=\frac{\sqrt{10}}3{}$​​​​​​​

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MCQ 281 Mark
If $\tan\theta=\frac{8}{15}$ then $\text{cosec}\ \theta=?$
  • ✓
    $\frac{17}{8}$
  • B
    $\frac{8}{17}$
  • C
    $\frac{17}{15}$
  • D
    $\frac{15}{17}$
Answer
Correct option: A.
$\frac{17}{8}$
Consider $\triangle\text{ABC}$
where $\angle\text{B}=90^\circ,\angle\text{A}=\theta.$
Then, $\tan\theta=\frac{\text{perpendicular}}{\text{Base}}$
$=\frac{\text{BC}}{\text{AB}}=\frac{8}{15}$
Let $\text{BC}={8}\text{k}$ and $\text{AB}=15\text{k},$ where $k$ is positive.
By Pythagoras Theorem,
$\text{AC}^2=\text{AB}^2+\text{BC}^2$
$\Rightarrow\text{AC}^2=(15\text{k})^2+(8\text{k})^2=225\text{k}^2+64\text{k}^2=289\text{k}^2$
$\Rightarrow\text{AC}=\text{17k}$
Now, $\text{cosec}\theta=\frac{\text{Hypotenuse}}{\text{Perpendicular}}$
$\frac{\text{AC}}{\text{BC}}=\frac{17\text{k}}{8\text{k}}=\frac{17}{8}$
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MCQ 291 Mark
$(\sec\text{A}+\tan\text{A})(1-\sin\text{A})=?$
  • A
    $\sin\text{A}$
  • ✓
    $\cos\text{A}$
  • C
    $\sec\text{A}$
  • D
    $\text{cosec}\text{A}$
Answer
Correct option: B.
$\cos\text{A}$
$(\sec\text{A}+\tan\text{A})(1-\sin\text{A})$
$=\Big(\frac{1}{\cos\text{A}}+\frac{\sin\text{A}}{\cos\text{A}}\Big)(1-\sin\text{A})$
$=\frac{(1+\sin\text{A})}{\cos\text{A}}\times(1-\sin\text{A})$
$=\frac{1-\sin^2\text{A}}{\cos\text{A}}$
$=\frac{\cos^2\text{A}}{\cos\text{A}}$
$=\cos\text{A}$
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MCQ 301 Mark
$\tan5^\circ\tan25^\circ\tan30^\circ\tan65^\circ\tan85^\circ=?$
  • A
    $\sqrt{3}$
  • ✓
    $\frac{1}{\sqrt{3}}$
  • C
    $1$
  • D
    None of these.
Answer
Correct option: B.
$\frac{1}{\sqrt{3}}$
$\tan5^\circ\tan25^\circ\tan30^\circ\tan65^\circ\tan85^\circ$
$=\tan5^\circ\times\tan25^\circ\times\frac{1}{\sqrt{3}}\times\tan(90^\circ-25^\circ)\times\tan(90^\circ-5^\circ)$
$=\tan5^\circ\times\tan25^\circ\times\frac{1}{\sqrt{3}}\times\cot25^\circ\times\cot5^\circ$
$=\tan5^\circ\cot5^\circ\tan25^\circ\cot25^\circ\times\frac{1}{\sqrt{3}}$
$=1\times1\times\frac{1}{\sqrt{3}}$
$=\frac{1}{\sqrt{3}}$
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MCQ 311 Mark
If $\sin\theta=\frac{1}{2}$ then $\cot\theta=?$
  • A
    $\frac{1}{\sqrt3}$
  • ✓
    $\sqrt{3}$
  • C
    $\frac{\sqrt3}{2}$
  • D
    $1$
Answer
Correct option: B.
$\sqrt{3}$
$\sin^2\theta+\cos^2\theta=1$
$\Rightarrow\Big(\frac{1}2{}\Big)^2+\cos^2\theta=1$
$\Rightarrow\cos^2\theta=1-\frac{1}{4}=\frac{3}{4}$
$\Rightarrow\cos\theta=\frac{\sqrt3}{2}$
$\therefore\cot\theta=\frac{\cos\theta}{\sin\theta}$
$=\frac{\frac{\sqrt3}{2}}{\frac{1}{2}}$
$=\sqrt3$
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MCQ 321 Mark
$\cot1^\circ\cos2^\circ\cos3^\circ\dots\cos180^\circ=?$
  • A
    $-1$
  • B
    $1$
  • ✓
    $0$
  • D
    $\frac12$
Answer
Correct option: C.
$0$
Since $\cos90^\circ=0$
$\cos1^\circ\cos2^\circ\cos3^\circ\dots\cos90^\circ\dots\cos180^\circ=0$
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MCQ 331 Mark
$\sqrt{\frac{1+\cos\text{A}}{1-\cos\text{A}}}=?$
  • A
    $\text{cosec }\text{A}-\cot\text{A}$
  • ✓
    $\text{cosec }\text{A}+\cot\text{A}$
  • C
    $\text{cosec }\text{A}\cot\text{A}$
  • D
    None of these.
Answer
Correct option: B.
$\text{cosec }\text{A}+\cot\text{A}$
$\sqrt{\frac{1+\cos\text{A}}{1-\cos\text{A}}}=\sqrt{\frac{1+\cos\text{A}}{1-\cos\text{A}}\times\frac{1+\cos\text{A}}{1+\cos\text{A}}}$
$=\sqrt{\frac{(1+\cos\text{A})^2}{1-\cos^2\text{A}}}$
$=\sqrt{\frac{(1+\cos\text{A})^2}{\sin^2\text{A}}}$
$=\frac{1+\cos\text{A}}{\sin\text{A}}$
$=\frac{1}{\sin\text{A}}-\frac{\cos\text{A}}{\sin\text{A}}$
$=\text{cosec }\text{A}+\cot\text{A}$
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MCQ 341 Mark
$(\text{cosec }\theta-\cot\theta)^2=?$
  • A
    $\frac{1+\cos\theta}{1-\cos\theta}$
  • ✓
    $\frac{1-\cos\theta}{1+\cos\theta}$
  • C
    $\frac{1+\sin\theta}{1-\sin\theta}$
  • D
    None of these.
Answer
Correct option: B.
$\frac{1-\cos\theta}{1+\cos\theta}$
$(\text{cosec }\theta-\cot\theta)^2=\Big(\frac{1}{\sin\theta}-\frac{\cos\theta}{\sin\theta}\Big) ^2$
$=\Big(\frac{1-\cos\theta}{\sin\theta}\Big)^2$
$=\frac{(1-\cos\theta)^2}{\sin^2\theta}$
$=\frac{(1-\cos\theta)^2}{1-\cos^2\theta}$
$=\frac{(1-\cos\theta)^2}{(1-\cos\theta)(1+\cos\theta)}$
$=\frac{1-\cos\theta}{1+\cos\theta}$
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MCQ 351 Mark
If $\tan\theta=\frac{4}{3}$ then $(\sin\theta+\cos\theta)=?$
  • A
    $\frac{7}{3}$
  • B
    $\frac{7}{4}$
  • ✓
    $\frac{7}{5}$
  • D
    $\frac{5}{7}$
Answer
Correct option: C.
$\frac{7}{5}$
Consider $\triangle\text{ABC}$
where $\angle\text{A}=90^\circ,\angle\text{A}=\theta$
Then, $\tan\theta=\frac{\text{Perpendicular}}{\text{Base}}=\frac{\text{BC}}{\text{AB}}=\frac{4}{3}$
Let $BC = 4k$ and $AB = 3k,$
where $k$ is positive.
By pythagoras theorem,
$\text{AC}^2=\text{AB}^2+\text{BC}^2$
$\Rightarrow\text{AC}^2=(\text{3k})^2+(\text{4k})^2=\text{9k}^2+\text{16k}^2=\text{25k}^2$
$\Rightarrow\text{AC}=\text{5k}$
Now, $\sin\theta=\frac{\text{Perpendicular}}{\text{Hypotenuse}}=\frac{\text{BC}}{\text{AC}}=\frac{\text{4k}}{\text{5k}}=\frac{4}{5}$
And, $\cos\theta=\frac{\text{Base}}{\text{Hypotenuse}}=\frac{\text{AB}}{\text{AC}}=\frac{3\text{k}}{\text{5k}}=\frac{3}{5}$
$\therefore\sin\theta+\cos\theta=\frac{4}{5}+\frac{3}{5}=\frac{7}{5}$
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MCQ 361 Mark
$\text{cosec}^257^\circ-\tan^233^\circ=?$
  • A
    $0$
  • ✓
    $1$
  • C
    $-1$
  • D
    $2$
Answer
Correct option: B.
$1$
$\text{cosec}^257^\circ-\tan^233^\circ$
$=\text{cosec}^2(90^\circ-33^\circ)-\tan^233^\circ$
$=\sec^230^\circ-\tan^233^\circ$
$=1$
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MCQ 371 Mark
$\sin47^\circ\cos43^\circ+\cos47^\circ\sin43^\circ=?$
  • A
    $\sin4^\circ$
  • B
    $\cos4^\circ$
  • ✓
    $1$
  • D
    $0$
Answer
Correct option: C.
$1$
$\sin47^\circ\cos43^\circ+\cos47^\circ\sin43^\circ$
$=\sin47^\circ\cos(90^\circ-47^\circ)+\cos47^\circ\sin(90^\circ-47^\circ)$
$=\sin47^\circ\sin47^\circ+\cos47^\circ\cos47^\circ$
$=\sin^247^\circ+\cos^247^\circ$
$=1$
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MCQ 381 Mark
If $\sin\text{A}+\sin^2\text{A}=1$ then $\cos^2\text{A}+\cos^4\text{A}=?$
  • A
    $ \frac{1}{2}$
  • ✓
    $1$
  • C
    $2$
  • D
    $3$
Answer
Correct option: B.
$1$
$\sin\text{A}+\sin^2\text{A}=1$
$\Rightarrow\sin\text{A}=1-\sin^2\text{A}$
$\Rightarrow\sin\text{A}=\cos^2\text{A}$
Now, $\cos^2\text{A}+\cos^4\text{A}=\sin\text{A}+\sin^2\text{A}=1$
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MCQ 391 Mark
$\frac{\sec30^\circ}{\text{cosec }60^\circ}=?$
  • A
    $\frac{2}{\sqrt{3}}$
  • B
    $\frac{\sqrt{3}}{2}$
  • C
    $\sqrt{3}$
  • ✓
    $1$
Answer
Correct option: D.
$1$
$\frac{\sec30^\circ}{\text{cosec }60^\circ}=\frac{\frac{2}{\sqrt{3}}}{\frac{2}{\sqrt{3}}}=1$
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MCQ 401 Mark
If $\cos\text{A}+\cos^2\text{A}=1$ then $\big(\sin^2\text{A}+\sin^4\text{A}\big)=?$
  • A
    $\frac{1}2{}$
  • B
    $2$
  • ✓
    $1$
  • D
    $4$
Answer
Correct option: C.
$1$
$\cos\text{A}+\cos^2\text{A}=1$
$\Rightarrow\cos\text{A}=\sin^2\text{A}\dots(\text{i})$
Squaring both sides of $(i),$ we get:
$\cos^2\text{A}=\sin^4\text{A}\dots(\text{ii})$
Adding $(i)$ and $(ii)$, we get:
$\sin^2\text{A}=\sin^4\text{A}=\cos\text{A}+\cos^2\text{A}$
$\Rightarrow\sin^2\text{A}+\sin^4\text{A}=1\\ [\because\cos\text{A}+\cos^2\text{A}=1]$
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MCQ 411 Mark
If $\tan^245^\circ-\cos^230^\circ=\text{x}\sin45^\circ\cos45^\circ$ then $\text{x}=?$
  • A
    $2$
  • B
    $-2$
  • ✓
    $\frac{1}{2}$
  • D
    $\frac{-1}{2}$
Answer
Correct option: C.
$\frac{1}{2}$
$\tan^245^\circ-\cos^230^\circ=\text{x}\sin45^\circ\cos45^\circ$
$\Rightarrow(1)^2-\Big(\frac{\sqrt3}{2}\Big)^2=\text{x}\times\frac{1}{\sqrt2}\times\frac{1}{\sqrt2}$
$\Rightarrow1-\frac{3}{4}=\text{x}\times\frac{1}{2}$
$\Rightarrow\frac{1}{4}=\frac{\text{x}}2{}$
$\Rightarrow\text{x}=\frac{1}{2}$
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MCQ 421 Mark
If $\text{3x}=\text{cosec}\theta$ and $\frac{3}{\text{x}}=\cot\theta$ then $3\Big(\text{x}^2-\frac{1}{\text{x}^2}\Big)=?$
  • A
    $\frac{1}{27}$
  • B
    $\frac{1}{81}$
  • ✓
    $\frac{1}{3}$
  • D
    $\frac{1}{9}$
Answer
Correct option: C.
$\frac{1}{3}$
We know that
$\text{cosec}^2\theta-\cot^2\theta=1$
$\Rightarrow(\text{3x})^2-\Big(\frac{3}{\text{x}}\Big)^2=1$
$\Rightarrow\text{9x}^2-\frac{9}{\text{x}^2}=1$
$\Rightarrow9\Big(\text{x}^2-\frac{1}{\text{x}^2}\Big)=1$
$\Rightarrow\Big(\text{x}^2-\frac{1}{\text{x}^2}\Big)=\frac{1}9{}$
$\Rightarrow3\Big(\text{x}^2-\frac{1}{\text{x}^2}\Big)=\frac{1}3{}$
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MCQ 431 Mark
$\cos^230^\circ\cot^245^\circ+4\sec^260^\circ+\frac{1}{2}\cos^290^\circ-2\tan^260^\circ=?$
  • A
    $\frac{73}{8}$
  • B
    $\frac{75}{8}$
  • C
    $\frac{81}{8}$
  • ✓
    $\frac{83}{8}$
Answer
Correct option: D.
$\frac{83}{8}$
$\cos^230^\circ\cot^245^\circ+4\sec^260^\circ+\frac{1}{2}\cos^290^\circ-2\tan^260^\circ$
$=\Big(\frac{\sqrt3}{2}\Big)^2\times\Big(\frac{1}{\sqrt2}\Big)^2+4(2)^2+\frac{1}{2}\times0-2\big(\sqrt3\big)^2$
$=\frac{3}{4}\times\frac{1}{2}+16+0-6$
$=\frac{3}{8}+10$
$=\frac{3+80}{8}$
$=\frac{83}{8}$
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MCQ 441 Mark
$\sqrt{\frac{1-\sin\text{A}}{1+\sin\text{A}}}=?$
  • A
    $\sec\text{A}+\tan\text{A}$
  • ✓
    $\sec\text{A}-\tan\text{A}$
  • C
    $\sec\text{A}\tan\text{A}$
  • D
    None of these.
Answer
Correct option: B.
$\sec\text{A}-\tan\text{A}$
$\sqrt{\frac{1-\sin\text{A}}{1+\sin\text{A}}}=\sqrt{\frac{1-\sin\text{A}}{1+\sin\text{A}}\times\frac{1-\sin\text{A}}{1-\sin\text{A}}}$
$=\sqrt{\frac{(1-\sin\text{A})^2}{1-\sin^2\text{A}}}$
$=\sqrt{\frac{(1-\sin\text{A})^2}{\cos^2\text{A}}}$
$=\frac{1-\sin\text{A}}{\cos\text{A}}$
$=\frac{1}{\cos\text{A}}-\frac{\sin\text{A}}{\cos\text{A}}$
$=\sec\text{A}-\tan\text{A}$
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MCQ 451 Mark
$(\cos0^\circ+\sin30^\circ+\sin45^\circ)(\sin90^\circ+\cos60^\circ-\cos45^\circ)=?$
  • A
    $\frac{5}{6}$
  • B
    $\frac{5}{8}$
  • C
    $\frac{3}{5}$
  • ✓
    $\frac{7}{4}$
Answer
Correct option: D.
$\frac{7}{4}$
$(\cos0^\circ+\sin30^\circ+\sin45^\circ)(\sin90^\circ+\cos60^\circ-\cos45^\circ)$
$=\Big(1+\frac{1}{2}+\frac{1}{\sqrt2}\Big)\Big(1+\frac{1}{2}-\frac{1}{\sqrt2}\Big)$
$=\Big(\frac{3}{2}+\frac{1}{\sqrt2}\Big)\Big(\frac{3}{2}-\frac{1}{\sqrt2}\Big)$
$=\Big(\frac{3}{2}\Big)^2-\Big(\frac{1}{\sqrt2}\Big)^2$
$=\frac{9}{4}-\frac{1}{2}$
$=\frac{9-2}{4}$
$=\frac{7}{4}$
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MCQ 461 Mark
If $\tan\text{x}-3\cot\text{x}$ then $\text{x}=?$
  • A
    $45^\circ$
  • ✓
    $60^\circ$
  • C
    $30^\circ$
  • D
    $15^\circ$
Answer
Correct option: B.
$60^\circ$
$\tan\text{x}-3\cot\text{x}$
$\Rightarrow\tan\text{x}=3\times\frac{1}{\tan\text{x}}$
$\Rightarrow\tan^2\text{x}=3$
$\Rightarrow\tan\text{x}=\sqrt3$
$\Rightarrow\tan\text{x}=\tan60^\circ$
$\Rightarrow\text{x}=60^\circ$
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MCQ 471 Mark
If $\tan\theta=\frac{\text{a}}{\text{b}}$ then $\frac{(\text{a}\sin\theta-\text{b}\cos\theta)}{(\text{a}\sin\theta+\text{b}\cos\theta)} =?$
  • A
    $\frac{(\text{a}^2+\text{b}^2)}{(\text{a}^2-\text{b}^2)}$
  • ✓
    $\frac{(\text{a}^2-\text{b}^2)}{(\text{a}^2+\text{b}^2)}$
  • C
    $\frac{\text{a}^2}{(\text{a}^2+\text{b}^2)}$
  • D
    $\frac{\text{b}^2}{(\text{a}^2+\text{b}^2)}$
Answer
Correct option: B.
$\frac{(\text{a}^2-\text{b}^2)}{(\text{a}^2+\text{b}^2)}$
Given, $\tan\theta=\frac{\text{a}}{\text{b}}$
Now, $\frac{(\text{a}\sin\theta-\text{b}\cos\theta)}{(\text{a}\sin\theta+\text{b}\cos\theta)} $
$=\frac{\frac{\text{a}\sin\theta}{\cos\theta}-\frac{\text{b}\cos\theta}{\cos\theta}}{\frac{\text{a}\sin\theta}{\cos\theta}+\frac{\text{b}\cos\theta}{\cos\theta}}$
$=\frac{\text{a}\tan\theta-\text{b}}{\text{a}\tan\theta+\text{b}}$
$=\frac{\text{a}\times\frac{\text{a}}{\text{b}}-\text{b}}{\text{a}\times\frac{\text{a}}{\text{b}}+\text{b}}$
$=\frac{\frac{\text{a}^2-\text{b}^2}{\text{b}}}{\frac{\text{a}^2+\text{b}^2}{\text{b}}}$
$=\frac{\text{a}^2-\text{b}^2}{\text{a}^2+\text{b}^2}$
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MCQ 481 Mark
If $(\tan\theta+\cot\theta)=5$ then $(\tan^2\theta+\cot^2\theta)=?$
  • A
    $27$
  • B
    $25$
  • C
    $24$
  • ✓
    $23.$
Answer
Correct option: D.
$23.$
$(\tan\theta+\cot\theta)=5$
$\Rightarrow(\tan\theta+\cot\theta)^2=5^2$
$\Rightarrow\tan^2\theta+\cot^2\theta+2\tan\theta\cot\theta=25$
$\Rightarrow\tan^2\theta+\cot^2\theta+2(1)=25$
$\Rightarrow\tan^2\theta+\cot^2\theta=23$
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MCQ 491 Mark
If $2\sin2\theta=\sqrt3$ then $\theta=?$
  • ✓
    $30^\circ$
  • B
    $45^\circ$
  • C
    $60^\circ$
  • D
    $90^\circ$
Answer
Correct option: A.
$30^\circ$
$2\cos2\theta=\sqrt3$
$\Rightarrow\sin2\theta=\frac{\sqrt3}{2}$
$\Rightarrow\sin2\theta=\sin60^\circ$
$\Rightarrow2\theta=60^\circ$
$\Rightarrow\theta=30^\circ$
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MCQ 501 Mark
$\sin(45^\circ+\theta)-\cos(45^\circ-\theta)=?$
  • A
    $2\sin\theta$
  • B
    $2\cos\theta$
  • ✓
    $0$
  • D
    $1$
Answer
Correct option: C.
$0$

$\sin(45^\circ+\theta)-\cos(45^\circ-\theta)$
$=\sin(45^\circ+\theta)-\cos[90^\circ-(45^\circ+\theta)]$
$=\sin(45^\circ+\theta)-\sin(45^\circ+\theta)$
$=0$

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MCQ 511 Mark
$\frac{2\tan^230^\circ\sec^252^\circ\sin^238^\circ}{\text{cosec}^270^\circ-\tan^220^\circ}=?$
  • A
    $2$
  • B
    $\frac{1}{2}$
  • ✓
    $\frac{2}{3}$
  • D
    $\frac{3}{2}$
Answer
Correct option: C.
$\frac{2}{3}$
$\frac{2\tan^230^\circ\sec^252^\circ\sin^238^\circ}{\text{cosec}^270^\circ-\tan^220^\circ}$
$=\frac{2\times\big(\frac{1}{\sqrt3}\big)^2\times\frac{1}{\cos^252^\circ}\times\sin^2(90^\circ-52^\circ)}{\text{cosec}^2(90^\circ-20^\circ)-\tan^220^\circ}$
$=\frac{2\times\frac{1}{3}\times\frac{1}{\cos^252^\circ}\times\cos^252^\circ}{\sec^220^\circ-\tan^220^\circ}$
$=\frac{\frac{2}{3}}1{}$
$=\frac{2}{3}$
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MCQ 521 Mark
$\sin^230^\circ+4\cot^245^\circ-\sec^260^\circ=?$
  • A
    $0$
  • ✓
    $\frac{1}{4}$
  • C
    $4$
  • D
    $1$
Answer
Correct option: B.
$\frac{1}{4}$
$\sin^230^\circ+4\cot^245^\circ-\sec^260^\circ$
$=\Big(\frac{1}{2}\Big)^2+4(1)^2-(2)^2$
$=\frac{1}{4}+4-4$
$=\frac{1}4{}$
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MCQ 531 Mark
If $\tan\theta=\frac{1}{\sqrt7}$ then $\frac{(\text{cosec}^2\theta-\sec^2\theta)}{(\text{cosec}^2\theta+\sec^2\theta)}=? $
  • A
    $\frac{-2}{3}$
  • B
    $\frac{-3}{4}$
  • C
    $\frac{2}{3}$
  • ✓
    $\frac{3}{4}$
Answer
Correct option: D.
$\frac{3}{4}$
$\tan\theta=\frac{1}{\sqrt7}$
$\Rightarrow\cot\theta=\frac{1}{\tan\theta}=\sqrt7$
Now, $\sec^2\theta=(1+\tan^2\theta)$
$1+\Big(\frac{1}{\sqrt7}\Big)^2=1+\frac{1}{7}=\frac{8}{7}$
And, $\text{cosec}^2\theta=(1+\cot^2\theta)$
$=1+\big(\sqrt7\big)^2=1+7=8$
$\therefore\frac{(\text{cosec}^2\theta-\sec^2\theta)}{(\text{cosec}^2\theta+\sec^2\theta)}=\frac{8-\frac{8}{7}}{8+\frac{8}{7}}$
$=\frac{\frac{56-8}{7}}{\frac{56+8}{7}}$
$=\frac{48}{64}$
$=\frac{3}{4}$
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MCQ 541 Mark
If $\cos\text{A}+\cos^2\text{A}=1$ then $\sin^2\text{A}+\sin^4\text{A}=?$
  • ✓
    $1$
  • B
    $2$
  • C
    $4$
  • D
    $3.$
Answer
Correct option: A.
$1$
$\cos\text{A}+\cos^2\text{A}=1$
$\Rightarrow\cos\text{A}=1-\cos^2\text{A}$
$\Rightarrow\cos\text{A}=\sin^2\text{A}$
Now, $\sin^2\text{A}+\sin^4\text{A}=\cos\text{A}+\cos^2\text{A}=1$
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MCQ 551 Mark
If $A$ and $B$ are acute angles such that $\sin\text{A}=\cos\text{B}$ then $(\text{A}+\text{B})=?$
  • A
    $45^\circ$
  • B
    $60^\circ$
  • ✓
    $90^\circ$
  • D
    $180^\circ$
Answer
Correct option: C.
$90^\circ$

$\sin\text{A}=\cos\text{B}$
$\Rightarrow\sin\text{A}=\sin(90^\circ-\text{B})$
$\Rightarrow\text{A}=90^\circ-\text{B}$
$\Rightarrow\text{A}+\text{B}=90^\circ$

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MCQ 561 Mark
$\frac{2\sin^263^\circ+1+2\sin^227^\circ}{3\cos^217^\circ-2+3\cos^273^\circ}=?$
  • A
    $\frac32$
  • B
    $\frac23$
  • C
    $2$
  • ✓
    $3$
Answer
Correct option: D.
$3$
$\frac{2\sin^263^\circ+1+2\sin^227^\circ}{3\cos^217^\circ-2+3\cos^273^\circ}$
$=\frac{2\sin^263^\circ+2\sin^227+1}{3\cos^217^\circ+3\cos^273^\circ-2}$
$=\frac{2\sin^263^\circ+2\cos^263^\circ+1}{3\cos^217^\circ+3\sin^217^\circ-2}$
$=\frac{2\times1+1}{3\times1-2}$
$=\frac{2+1}{3-2}$
$=3$
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MCQ 571 Mark
If $\text{2x}=\sec\text{A}$ and $\frac{2}{\text{x}}=\tan\text{A}$ then $2\Big(\text{x}^2-\frac{1}{\text{x}^2}\Big)=?$
  • ✓
    $\frac{1}{2}$
  • B
    $\frac{1}{4}$
  • C
    $\frac{1}{8}$
  • D
    $\frac{1}{16}$
Answer
Correct option: A.
$\frac{1}{2}$
We know that
$\sec^2\text{A}-\tan^2\text{A}=1$
$\Rightarrow(\text{2x})^2-\Big(\frac{2}{\text{x}}\Big)^2=1$
$\Rightarrow\text{4x}^2-\frac{4}{\text{x}^2}=1$
$\Rightarrow4\Big(\text{x}^2-\frac{1}{\text{x}^2}\Big)=1$
$\Rightarrow\Big(\text{x}^2-\frac{1}{\text{x}^2}\Big)=\frac{1}4{}$
$\Rightarrow2\Big(\text{x}^2-\frac{1}{\text{x}^2}\Big)=\frac{1}2{}$
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