Question

Find the value of x for which
(i) x tan 45° cot 60° = sin 30° cosec 60°
(ii) $2{\mathrm{cosec}}^{2}30°+x{\sin }^{2}60°-\frac{3}{4}{\tan }^{2}30°=10$

Answer 1

(i) As we know that,
​$$\begin{gathered} \tan 45°=1 \\ \cot 60°=\frac{1}{\sqrt{3}} \\ \sin 30°=\frac{1}{2} \\ \mathrm{cosec}60°=\frac{2}{\sqrt{3}} \\ \mathrm{On}\mathrm{substituting}\mathrm{these}\mathrm{values},\mathrm{we}\mathrm{get} \\ x\tan 45°\cot 60°=\sin 30°\mathrm{cosec}60° \\ \Rightarrow x\left(1\right)\left(\frac{1}{\sqrt{3}}\right)=\left(\frac{1}{2}\right)\left(\frac{2}{\sqrt{3}}\right) \\ \Rightarrow \frac{x}{\sqrt{3}}=\frac{1}{\sqrt{3}} \\ \Rightarrow x=1 \\ \mathrm{Hence},\mathrm{the}\mathrm{value}\mathrm{of}x\mathrm{is}1. \end{gathered}$$


(ii) As we know that,
​$$\begin{gathered} \tan 30°=\frac{1}{\sqrt{3}} \\ \sin 60°=\frac{\sqrt{3}}{2} \\ \mathrm{cosec}30°=2 \\ \mathrm{On}\mathrm{substituting}\mathrm{these}\mathrm{values},\mathrm{we}\mathrm{get} \\ 2{\mathrm{cosec}}^{2}30°+x{\sin }^{2}60°-\frac{3}{4}{\tan }^{2}30°=10 \\ \Rightarrow 2{\left(2\right)}^2+x{\left(\frac{\sqrt{3}}{2}\right)}^2-\frac{3}{4}{\left(\frac{1}{\sqrt{3}}\right)}^2=10 \\ \Rightarrow 2\left(4\right)+x\left(\frac{3}{4}\right)-\frac{3}{4}\left(\frac{1}{3}\right)=10 \\ \Rightarrow 8+\frac{3x}{4}-\frac{1}{4}=10 \\ \Rightarrow \frac{3x}{4}-\frac{1}{4}=10-8 \\ \Rightarrow \frac{3x-1}{4}=2 \\ \Rightarrow 3x-1=8 \\ \Rightarrow 3x=9 \\ \Rightarrow x=3 \\ \mathrm{Hence},\mathrm{the}\mathrm{value}\mathrm{of}x\mathrm{is}3. \end{gathered}$$