[2 Mark Question Answer]

Question 12 Marks

If $3\cos\theta - 4\sin\theta = 2\cos\theta + \sin\theta ,$ find $\tan\theta .$

Answer

$3 \cos \theta-4 \sin \theta=2 \cos \theta+\sin \theta$
$ \Rightarrow 3 \cos \theta-2 \cos \theta=\sin \theta+4 \sin \theta $
$ \Rightarrow \cos \theta=5 \sin \theta $
$ \Rightarrow \frac{\sin \theta}{\cos \theta}=\frac{1}{5}$
$\Rightarrow \tan \theta=\frac{1}{5}$

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Question 22 Marks

In the given figure, $\text{PQR}$ is a triangle, in which $\text{QS} \perp \text{PR}, \text{QS} = 3\ cm, \text{PS} = 4\ cm$ and $\text{QR} = 12\ cm,$ find the value of: $\cot^2P -\operatorname{cosec}^2P$

Answer

$\cos P=\frac{PS}{P Q}=\frac{4}{5}$
$\cot ^2 P-\operatorname{cosec}^2 P$
$=\left(\frac{\cos P}{\sin P}\right)^2-\left(\frac{1}{\sin P}\right)^2$
$=\left(\frac{\frac{4}{5}}{\frac{3}{5}}\right)^2-\left(\frac{1}{\frac{3}{5}}\right)^2$
$=\left(\frac{4}{3}\right)^2-\left(\frac{5}{3}\right)^2$
$=\frac{16}{9}-\frac{25}{9}$
$=-\frac{9}{9}$
$=-1 .$

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Question 32 Marks

In the given figure, ${PQR}$ is a triangle, in which $\text{QS} \perp \text{PR}, \text{QS} = 3\ cm, \text{PS} = 4\ cm$ and $\text{QR} = 12\ cm,$ find the value of $:\sin P$

Answer

$\triangle QSP$ is a right$-$angled triangle.
$ \therefore \text{PQ}^2=\text{QS}^2+\text{PS}^2$
$=3^2+4^2$
$=9+16$
$=25$
$\Rightarrow \text{PQ}=5 \ cm$
$\sin P$
$=\frac{QS}{PQ}$
$=\frac{3}{5} . $

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MATHEMATICS [2 Mark Question Answer]

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