1 Marks Question

Question 11 Mark

If $\sin \alpha=\frac{1}{2}$, then find the value of $3 \sin \alpha-4 \sin ^3 \alpha$.

Answer

It is given that $\sin \alpha=\frac{1}{2}$,
So, substituting value of $\sin a$,
$3 \sin \alpha-4 \sin ^3 \alpha$
$=3\left(\frac{1}{2}\right)-4\left(\frac{1}{2}\right)^3$
$=\frac{3}{2}-4\left(\frac{1}{8}\right)$
$=\frac{3}{2}-\frac{1}{2}$
$=1$

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Question 21 Mark

If $\tan 2 A=\cot \left(A+60^{\circ}\right)$, find the value of $A$, where $2 A$ is an acute angle.

Answer

Given $\tan 2 A=\cot \left(A+60^{\circ}\right)$
$\Rightarrow \tan 2 A=\cot \left(A+60^{\circ}\right)$
As we know $\tan 2 A=\cot \left(90^{\circ}-2 A\right)$
$\Rightarrow \cot \left(90^{\circ}-2 A\right)=\cot \left(A+60^{\circ}\right)$
On equating the angles
$\Rightarrow 90^{\circ}-2 A=A+60^{\circ}$
$\Rightarrow 3 A=30^{\circ}$
Divide the above equation by $3$,
$\Rightarrow A=10^{\circ}$
Hence, the value of $A$ is $10^{\circ}$

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Question 31 Mark

Find the value of $\left(\sec ^2 \theta-1\right) \cdot \cot ^2 \theta$

Answer

$\left(\sec ^2 \theta-1\right) \cdot \cot ^2 \theta$
Using the identity $\sec ^2 \theta-1=\tan ^2 \theta$, we get
$\left(\sec ^2 \theta-1\right) \cdot \cot ^2 \theta=\tan ^2 \theta \cdot \cot ^2 \theta$
$=\frac{\sin ^2 \theta}{\cos ^2 \theta} \cdot \frac{\cos ^2 \theta}{\sin ^2 \theta}=1$
$\left[\text { As } \tan \theta=\frac{\sin \theta}{\cos \theta} \text { and } \cot \theta=\frac{\cos \theta}{\sin \theta}\right]$

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Question 41 Mark

Express $\operatorname{cosec} 48^{\circ}+\tan 88^{\circ}$ in terms of t-ratio of angle 00 and $45^{\circ}$

Answer

$
\operatorname{cosec} 48^{\circ}+\tan 88^{\circ}
$
Using the identity
$
\operatorname{cosec} \theta=\sec (90-\theta) \text { and } \tan \theta=\cot (90-\theta),
$
we get
$
\operatorname{cosec} 48^{\circ}+\tan 88^{\circ}=\sec (90-48)+\cot (90-88)
$
Thus $\operatorname{cosec} 48^{\circ}+\tan 88^{\circ}=\sec 42^{\circ}+\tan 2^{\circ}$

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Question 51 Mark

Evaluate: $\sin ^2 A+\cos ^2 A+\cot ^2 A$

Answer

$\sin ^2 A+\cos ^2 A+\cot ^2 A ($Given$)$
We know that $\sin ^2 A+\cos ^2 A=1$
$\Rightarrow \sin ^2 A+\cos ^2 A+\cot ^2 A=1+\cot ^2 A$
$\Rightarrow \operatorname{cosec}^2 A\left(\text { Since } 1+\cot ^2 A=\operatorname{cosec}^2 A\right)$
Hence, $\sin ^2 A+\cos ^2 A+\cot ^2 A=\operatorname{cosec}^2 A$

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Question 61 Mark

Given $\sqrt{3} \tan 5 \theta=1$, find the value of $\theta$.

Answer

$\sqrt{3} \tan 5 \theta=1$
$\Rightarrow \tan 5 \theta=\frac{1}{\sqrt{3}}$
We know that $\tan 30^{\circ}=\frac{1}{\sqrt{3}}$
$\Rightarrow 5 \theta=30^{\circ}$
$\Rightarrow \theta=\frac{30^{\circ}}{5}=6^{\circ}$
Hence, the value of $\theta$ is $6^{\circ}$.

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Question 71 Mark

If $24 \cot A=7$, find the value of $\sin A$.

Answer

$24 \cot A=7$
$\Rightarrow \cot A=\frac{7}{24}$
$\cot A=\frac{\text { Base }}{\text { Perpendicular }}$
$\Rightarrow a=7$ and $b=24$
Using the Pythagoras theorem we get,
$c^2=a^2+b^2$
$\Rightarrow c^2=(7)^2+(24)^2$
$\Rightarrow c^2=49+576=625$
$\Rightarrow c=25$
We know that $\sin A=\frac{\text { Perpendicular }}{\text { hypotenuse }}$
$\Rightarrow \sin A=\frac{24}{25}$

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Question 81 Mark

Find the value of $\cos \theta+\sec \theta$, when it is given that $\cos \theta=\frac{1}{2}$

Answer

Since, $\cos \theta=\frac{1}{2}$
$
\therefore \sec \theta=\frac{1}{\cos \theta}=\frac{1}{\frac{1}{2}}=2
$
Hence,
$
\cos \theta+\sec \theta=\frac{1}{2}+2=\frac{1+4}{2}=\frac{5}{2}
$

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Question 91 Mark

If $\theta=45^{\circ}$, then find the value of $2 \operatorname{cosec}^2 \theta+3 \sec ^2 \theta$.

Answer

If $ \theta=45^{\circ} \text { then, }$
$\Rightarrow 2 \operatorname{cosec}^2 \theta+3 \sec ^2 \theta$
$=2 \operatorname{cosec}^2 45^{\circ}+3 \sec ^2 45^{\circ}$
$\Rightarrow 2(\sqrt{2})^2+3(\sqrt{2})^2$
$=2 \times 2+3 \times 2 $
$\Rightarrow 4+6 $
$\Rightarrow 10$
Hence the value of $2 \operatorname{cosec}^2 \theta+3 \sec ^2 \theta=10$

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Question 101 Mark

If $\sin A+\sin ^2 A=1$, then find the value of the expression $\left(\cos ^2 A+\cos ^4 A\right)$.

Answer

$\sin A +\sin ^2 A=1$
$\Rightarrow \sin A=1-\sin ^2 A$
But $ \cos ^2 A=1-\sin ^2 A$
$\Rightarrow \sin A =\cos ^2 A$
$\Rightarrow \sin ^2 A=\cos ^4 A$
$\cos ^2 A+\cos ^4 A$
$\Rightarrow \sin A +\sin ^2 A$
$\Rightarrow 1$

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Question 111 Mark

The value of $\frac{\tan 35^{\circ}}{\cot 55^{\circ}}+\frac{\cot 78^{\circ}}{\tan 12^{\circ}}$ is $............$

Answer

$\frac{\tan 35^{\circ}}{\cot 55^{\circ}}+\frac{\cot 78^{\circ}}{\tan 12^{\circ}}$
$\Rightarrow \frac{\tan 35^{\circ}}{\cot \left(90^{\circ}-35^{\circ}\right)}+\frac{\cot \left(90^{\circ}-12^{\circ}\right)}{\tan 12^{\circ}}$
$\Rightarrow \frac{\tan 35^{\circ}}{\tan 35^{\circ}}+\frac{\tan 12^{\circ}}{\tan 12^{\circ}}$
$\Rightarrow 1+1$
$\Rightarrow 2$

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Question 121 Mark

The value of $\sin 32^{\circ} \cos 58^{\circ}+\cos 32^{\circ} \sin 58^{\circ}$ is $.........$

Answer

$\sin 32^{\circ} \cos 58^{\circ}+\cos 32^{\circ} \sin 58^{\circ}$
$\Rightarrow \sin 32^{\circ} \cdot \cos \left(90^{\circ}-32^{\circ}\right)+\cos 32^{\circ} \cdot \sin \left(90^{\circ}-32^{\circ}\right)$
$\Rightarrow \sin 32^{\circ} \cdot \sin 32^{\circ}+\cos 32^{\circ} \cdot \cos 32^{\circ}$
$\Rightarrow \sin 32^{\circ}+\cos ^2 32^{\circ}$
$\Rightarrow 1$

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Question 131 Mark

If $\cos A=\sin 42^{\circ}$, then find the value of $A$.

Answer

$\cos A =\sin 42^{\circ}$
$\Rightarrow \cos A =\sin \left(90^{\circ}-48^{\circ}\right)$
$\Rightarrow \cos A =\cos 48^{\circ}$
$\Rightarrow A =48^{\circ}$

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Question 141 Mark

Evaluate: $\tan 40^{\circ} \times \tan 50^{\circ}$

Answer

$\tan 40^{\circ} \times \tan \left(90^{\circ}-40^{\circ}\right)$
$\Rightarrow \tan 40^{\circ} \times \cot 40^{\circ}$
$\Rightarrow \tan 40^{\circ} \times \frac{1}{\tan 40^{\circ}}$
$\Rightarrow 1$

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Question 151 Mark

Express $\left(\sin 67^{\circ}+\cos 75^{\circ}\right)$ in terms of trigonometric ratios of the angle between $0^{\circ}$ and $45^{\circ}$.

Answer

$\sin 67^{\circ}+\cos 75^{\circ}=\sin (90-23)^{\circ}+\cos (90-15)^{\circ}$
$=\cos 23^{\circ}+\sin 15^{\circ}$
$[\because \cos (90-\theta)=\sin \theta \ \sin (90-\theta)=\cos \theta$

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Question 161 Mark

Evaluate: $\frac{\tan 65^{\circ}}{\cot 25^{\circ}}$

Answer

$\frac{\tan 65^{\circ}}{\cot 25^{\circ}}=\frac{\tan (90-25)^{\circ}}{\cot 25^{\circ}}$
$\frac{\cot 25^{\circ}}{\cot 25^{\circ}}=1[\because \tan (90-\theta)=\cot \theta]$

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