MCQ

MCQ 11 Mark

The value of $\tan 1^{\circ} \cdot \tan 2^{\circ} \tan 3^{\circ}$ equal to

A
-1
✓
1
C
$\frac{\pi}{2}$
D
2

Answer

Correct option: B.

1
Explanation:
$\tan 1^{\circ} \tan 2^{\circ} \tan 3^{\circ} \ldots \tan 89^{\circ}$
$=\left(\tan 1^{\circ} \tan 89^{\circ}\right)\left(\tan 2^{\circ} \tan 88^{\circ}\right)$
$\ldots\left(\tan 44^{\circ} \tan 46^{\circ}\right) \tan 45^{\circ}$
$=\left(\tan 1^{\circ} \cot 1^{\circ}\right)\left(\tan 2^{\circ} \cot 2^{\circ}\right)$
$\ldots\left(\tan 44^{\circ} \cot 44^{\circ}\right) \cdot \tan 45^{\circ}$
$\left.\ldots \tan \left(\because 90^{\circ}-\theta\right)=\cot \theta\right]$
$=1 \times 1 \times 1 \times \ldots \times 1 \times \tan 45^{\circ}=1$
$\tan 1^{\circ} \tan 2^{\circ} \tan 3^{\circ} \ldots \tan 89^{\circ}$
$=\left(\tan 1^{\circ} \tan 89^{\circ}\right)\left(\tan 2^{\circ} \tan 88^{\circ}\right)$
$\ldots\left(\tan 44^{\circ} \tan 46^{\circ}\right) \tan 45^{\circ}$
$=\left(\tan 1^{\circ} \cot 1^{\circ}\right)\left(\tan 2^{\circ} \cot 2^{\circ}\right)$
$\ldots\left(\tan 44^{\circ} \cot 44^{\circ}\right) \cdot \tan 45^{\circ}$
$\left.\ldots \tan \left(\because 90^{\circ}-\theta\right)=\cot \theta\right]$
$=1 \times 1 \times 1 \times \ldots \times 1 \times \tan 45^{\circ}=1$

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MCQ 21 Mark

The cotangent of the angles $\frac{\pi}{3}, \frac{\pi}{4}$ and $\frac{\pi}{6}$ are in

A
A.P.
✓
G.P.
C
H.P.
D
Not in progression

Answer

Correct option: B.

G.P.

G.P.
Explanation:
$\begin{array}{ll}
& \cot \frac{\pi}{3}=\frac{1}{\sqrt{3}}, \cot \frac{\pi}{4}=1, \cot \frac{\pi}{6}=\sqrt{3} \\
\therefore \quad & \cot \frac{\pi}{3} \cot \frac{\pi}{6}=1=\left(\cot \frac{\pi}{4}\right)^2 \\
\therefore \quad & \cot \frac{\pi}{3}, \cot \frac{\pi}{4}, \cot \frac{\pi}{6} \text { are in G.P. }
\end{array}$

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MCQ 31 Mark

If $\operatorname{cosec} \theta-\cot \theta=q$, then the value of $\cot \theta$ is

A
$\frac{2 q}{1+q^2}$
B
$\frac{2 q}{1-q^2}$
✓
$\frac{1-\mathrm{q}^2}{2 \mathrm{q}}$
D
$\frac{1+q^2}{2 q}$

Answer

Correct option: C.

$\frac{1-\mathrm{q}^2}{2 \mathrm{q}}$

$\frac{1-\mathrm{q}^2}{2 \mathrm{q}}$
Explanation:
$\operatorname{cosec} \theta-\cot \theta=q \ldots . . \text { (i) } $
$operatorname{cosec}^2 \theta-\cot ^2 \theta=1 $
$\therefore(\operatorname{cosec} \theta+\cot \theta)(\operatorname{cosec} \theta-\cot \theta)=1$
$\therefore(\operatorname{cosec} \theta+\cot \theta) q=1 $
$\therefore \operatorname{cosec} \theta+\cot \theta=1 / q \ldots \ldots \text {.ii) }$
Subtracting (i) from (ii), we get
$2 \cot \theta=\frac{1}{q}-q$
$\therefore \cot \theta=\frac{1-q^2}{2 q}$

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MCQ 41 Mark

$1-\frac{\sin ^2 \theta}{1+\cos \theta}+\frac{1+\cos \theta}{\sin \theta}-\frac{\sin \theta}{1-\cos \theta}$ equals

A
$0$
B
$1$
C
$\sin \theta$
✓
$\cos \theta$

Answer

Correct option: D.

$\cos \theta$

$\cos \theta$
Explanation:
$1-\frac{\sin ^2 \theta}{1+\cos \theta}+\frac{1+\cos \theta}{\sin \theta}-\frac{\sin \theta}{1-\cos \theta}$
$=1-\frac{1-\cos ^2 \theta}{1+\cos \theta}+\frac{1-\cos ^2 \theta}{\sin \theta(1-\cos \theta)}-\frac{\sin \theta}{1-\cos \theta} $
$=1-(1-\cos \theta)+\frac{\sin \theta}{1-\cos \theta}-\frac{\sin \theta}{1-\cos \theta} $
$=\cos \theta$

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MCQ 51 Mark

If $\operatorname{cosec} \theta+\cot \theta=\frac{5}{2}$, then the value of $\tan \theta$ is

A
$\frac{14}{25}$
✓
$\frac{20}{21}$
C
$\frac{21}{20}$
D
$\frac{15}{16}$

Answer

Correct option: B.

$\frac{20}{21}$

$\frac{20}{21}$
Explanation:
$\operatorname{cosec} \theta+\cot \theta=\frac{5}{2} \ldots \ldots \ldots \ldots \ldots \text { (i) } $
$\operatorname{cosec}^2 \theta-\cot ^2 \theta=1 $
$\therefore(\operatorname{cosec} \theta+\cot \theta)(\operatorname{cosec} \theta-\cot \theta)=1$
$\therefore \frac{5}{2}(\operatorname{cosec} \theta-\cot \theta)=1$
$\therefore \operatorname{cosec} \theta-\cot \theta=\frac{2}{5} \ldots \text { (ii) }$
Subtracting (ii) from (i), we get
$2 \cot \theta=\frac{5}{2}-\frac{2}{5}=\frac{21}{10} $
$\therefore \cot \theta=\frac{21}{20} $
$\therefore \tan \theta=\frac{20}{21}$

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MCQ 61 Mark

If $\sec \theta=m$ and $\tan \theta=n$, then $\frac{1}{m}\left\{(m+n)+\frac{1}{(m+n)}\right\}$ is equal to

✓
$2$
B
$m n$
C
$2 \mathrm{~m}$
D
$2 n$

Answer

Correct option: A.

$2$

$2$
Explanation:
$\frac{1}{m}\left[(m+n)+\frac{1}{(m+n)}\right] $
$=\frac{1}{\sec \theta}\left[(\sec \theta+\tan \theta)+\frac{1}{(\sec \theta+\tan \theta)}\right]$
$=\frac{1}{\sec \theta}\left(\sec \theta+\tan \theta+\frac{\sec \theta-\tan \theta}{\sec ^2 \theta-\tan ^2 \theta}\right) $
$=\frac{1}{\sec \theta}\left(\sec \theta+\tan \theta+\frac{\sec \theta-\tan \theta}{1}\right)$
$=\frac{1}{\sec \theta}(2 \sec \theta)=2$

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MCQ 71 Mark

If $\theta=60^{\circ}$, then $\frac{1+\tan ^2 \theta}{2 \tan \theta}$ is equal to

A
$\frac{\sqrt{3}}{2}$
B
$\frac{2}{\sqrt{3}}$
C
$\frac{1}{\sqrt{3}}$
✓
$\sqrt{3}$

Answer

Correct option: D.

$\sqrt{3}$

$\frac{2}{\sqrt{3}}$
Explanation:
$\begin{aligned}
\frac{1+\tan ^2 \theta}{2 \tan \theta} & =\frac{1+\tan ^2 60^{\circ}}{2 \tan 60^{\circ}}\\
& =\frac{1+(\sqrt{3})^2}{2 \sqrt{3}} \\
& =\frac{1+3}{2 \sqrt{3}} \\
& =\frac{2}{\sqrt{3}}
\end{aligned}$

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MCQ 81 Mark

If $\alpha$ is a root of $25 \cos ^2 \theta+5 \cos \theta-12=0, \frac{\pi}{2}<\alpha<\pi$, then $\sin 2 \alpha$ is equal to

✓
$-\frac{24}{25}$
B
$-\frac{13}{18}$
C
$\frac{13}{18}$
D
$\frac{24}{25}$

Answer

Correct option: A.

$-\frac{24}{25}$

$-\frac{24}{25}$
Explanation:
$25 \cos ^2 \theta+5 \cos \theta-12=0 $
$\therefore(5 \cos \theta+4)(5 \cos \theta-3)=0 $
$\therefore \cos \theta=-\frac{4}{5} \text { or } \cos \theta=\frac{3}{5} $
$\text { Since } \frac{\pi}{2}<\alpha<\pi$
$\cos \alpha<0$
$\therefore \cos \alpha=-\frac{4}{5}$
$\sin ^2 \alpha=1-\cos ^2 \alpha=1-\frac{16}{25}=\frac{9}{25} $
$\therefore \sin \alpha= \pm \frac{3}{5} $
$\text { Since } \frac{\pi}{2}<\alpha<\pi \sin \alpha>0$
$\therefore \sin \alpha=3 / 5$
$\sin 2 \alpha=2 \sin \alpha \cos \alpha $
$=2\left(\frac{3}{5}\right)\left(\frac{-4}{5}\right)=-\frac{24}{25}$

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MCQ 91 Mark

$\frac{\tan \mathrm{A}}{1+\sec \mathrm{A}}+\frac{1+\sec \mathrm{A}}{\tan \mathrm{A}}$ is equal to

✓
$2 \operatorname{cosec} A$
B
$2 \sec A$
C
$2 \sin \mathrm{A}$
D
$2 \cos A$

Answer

Correct option: A.

$2 \operatorname{cosec} A$

$2 \operatorname{cosec} A$
Explanation:
$\frac{\tan A}{1+\sec A}+\frac{1+\sec A}{\tan A}$
$= \frac{\tan ^2 A+1+\sec ^2 \mathrm{~A}+2 \sec \mathrm{A}}{(1+\sec \mathrm{A}) \tan \mathrm{A}} $
$=\frac{\sec ^2 \mathrm{~A}+\sec ^2 \mathrm{~A}+2 \sec \mathrm{A}}{(1+\sec \mathrm{A}) \tan \mathrm{A}} \ldots\left[\because 1+\tan ^2 \mathrm{~A}=\sec ^2 \mathrm{~A}\right] $
$= \frac{2 \sec \mathrm{A}(\sec \mathrm{A}+1)}{(1+\sec \mathrm{A}) \tan \mathrm{A}}=\frac{2 \sec \mathrm{A}}{\tan \mathrm{A}} $
$= \frac{2}{\sin \mathrm{A}}=2 \operatorname{cosec} \mathrm{A}$

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MCQ 101 Mark

The value of the expression $\cos 1^{\circ} \cdot \cos 2^{\circ} \cdot \cos 3^{\circ} \ldots \cos 179^{\circ}=$

A
-1
✓
0
C
$\frac{1}{\sqrt{2}}$
D
1

Answer

Correct option: B.

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