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Model Paper 8 — Maths STD 11 Science — Question

CBSE BoardEnglish MediumSTD 11 ScienceMathsModel Paper 85 Marks
Question
Prove that: $\cot x+\cot \left(\frac{\pi}{3}+x\right)+\cot \left(\frac{2 \pi}{3}+x\right)=3 \cot 3 x$
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Answer

We have to prove $\cot x+\cot \left(\frac{\pi}{3}+x\right)+\cot \left(\frac{2 \pi}{3}+x\right)=3 \cot 3 x$.
$\text{LHS} =\cot x+\cot \left(\frac{\pi}{3}+x\right)+\cot \left(\frac{2 \pi}{3}+x\right)$
We know,
$\cot \left(\frac{2 \pi}{3}+x\right)=\cot \left(\pi-\left(\frac{\pi}{3}-x\right)\right)=-\cot \left(\frac{\pi}{3}-x\right) \ldots\left(\right.$ as $-\cot \theta=\cot \left(180^{\circ}-\theta\right)$
Hence the above LHS becomes
$=\cot x+\cot \left(\frac{\pi}{3}+x\right)-\cot \left(\frac{\pi}{3}-x\right)$
$=\frac{1}{\tan x}+\frac{1}{\tan \left(\frac{\pi}{3}+x\right)} \cdot \frac{1}{\tan \left(\frac{\pi}{3}-x\right)}$
$=\frac{1}{\tan x}+\left(\frac{1-\tan x \tan \frac{\pi}{3}}{\tan \frac{\pi}{3}+\tan x}\right)-\left(\frac{1+\tan x \tan \frac{\pi}{3}}{\tan \frac{x}{3}-\tan x}\right) \cdots\left[\because \tan (A+B)=\left(\frac{\tan A+\tan B}{1-\tan A \tan B}\right) \text { and } \tan (A-B)=\left(\frac{\tan A-\tan B}{1+\tan A \tan B}\right)\right]$
$=\frac{1}{\tan x}+\left(\frac{1-\sqrt{3} \tan x}{\sqrt{3}+\tan x}\right)-\left(\frac{1+\sqrt{3} \tan x}{\sqrt{3}-\tan x}\right)$
$=\frac{1}{\tan x}+\left(\frac{(1-\sqrt{3} \tan x)(\sqrt{3}-\tan x)-(1+\sqrt{3} \tan x)(\sqrt{3}+\tan x)}{(\sqrt{3}+\tan x)(\sqrt{3}-\tan x)}\right)$
$=\frac{1}{\tan x}+\left(\frac{\left(\sqrt{3}-\tan x-3 \tan x+\sqrt{3} \tan ^2 x\right)-\left(\sqrt{3}+3 \tan x+\tan x+\sqrt{3} \tan ^2 x\right)}{\left(3-\tan ^2 x\right)}\right)$
$=\frac{1}{\tan x}+\left(\frac{(0-4 \tan x-4 \tan x+0)}{\left(3-\tan ^2 x\right)}\right)$
$=\frac{1}{\tan x}-\left(\frac{8 \tan x}{\left(\left(3-\tan ^2 x\right)\right)}\right)$
$=\left(\frac{\left(3-\tan ^2 x\right)-8 \tan ^2 x}{\tan x\left(3-\tan ^2 x\right)}\right)=\left(\frac{\left(3-\tan ^2 x\right)-8 \tan ^2 x}{\tan x\left(3-\tan ^2 x\right)}\right)$
$=3\left(\frac{1-3 \tan ^2 x}{\left(3 \tan x-\tan ^3 x\right)}\right)$
$=3 \times \frac{1}{\tan 3 x} \ldots\left(\operatorname{as} \tan 3 x=\frac{3 \tan x-\tan ^3 x}{1-3 \tan ^2 x}\right)$
$= \cot 3x$
$\text{LHS = RHS}$
Hence proved.

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