# Find the value of x for which

Question:

Find the value of x for which
(i) x tan 45° cot 60° = sin 30° cosec 60

(ii) $2 \operatorname{cosec}^{2} 30^{\circ}+x \sin ^{2} 60^{\circ}-\frac{3}{4} \tan ^{2} 30^{\circ}=10$

Solution:

(i) As we know that,

$\tan 45^{\circ}=1$

$\cot 60^{\circ}=\frac{1}{\sqrt{3}}$

$\sin 30^{\circ}=\frac{1}{2}$

$\operatorname{cosec} 60^{\circ}=\frac{2}{\sqrt{3}}$

On substituting these values, we get

$x \tan 45^{\circ} \cot 60^{\circ}=\sin 30^{\circ} \operatorname{cosec} 60^{\circ}$

$\Rightarrow x(1)\left(\frac{1}{\sqrt{3}}\right)=\left(\frac{1}{2}\right)\left(\frac{2}{\sqrt{3}}\right)$

$\Rightarrow \frac{x}{\sqrt{3}}=\frac{1}{\sqrt{3}}$

$\Rightarrow x=1$

Hence, the value of $x$ is 1 .

(ii) As we know that,

$\tan 30^{\circ}=\frac{1}{\sqrt{3}}$

$\sin 60^{\circ}=\frac{\sqrt{3}}{2}$

$\operatorname{cosec} 30^{\circ}=2$

On substituting these values, we get

$2 \operatorname{cosec}^{2} 30^{\circ}+x \sin ^{2} 60^{\circ}-\frac{3}{4} \tan ^{2} 30^{\circ}=10$

$\Rightarrow 2(2)^{2}+x\left(\frac{\sqrt{3}}{2}\right)^{2}-\frac{3}{4}\left(\frac{1}{\sqrt{3}}\right)^{2}=10$

$\Rightarrow 2(4)+x\left(\frac{3}{4}\right)-\frac{3}{4}\left(\frac{1}{3}\right)=10$

$\Rightarrow 8+\frac{3 x}{4}-\frac{1}{4}=10$

$\Rightarrow \frac{3 x}{4}-\frac{1}{4}=10-8$

$\Rightarrow \frac{3 x-1}{4}=2$

$\Rightarrow 3 x-1=8$

$\Rightarrow 3 x=9$

$\Rightarrow x=3$

Hence, the value of $x$ is 3 .