Question

$cotx-\tan x=$

• A. $cot2x$
• B. $2cot2x$
• C. $2cot2x$
• D. $co{t}^2\left(2x\right)$

Answer 1

The correct option is C

$2cot2x$

Explanation for the correct option:

Let,$\cot \left(x\right)-\tan \left(x\right)=\frac{\cos \left(x\right)}{\sin \left(x\right)}-\frac{\sin \left(x\right)}{\cos \left(x\right)}$

$=\frac{{\cos }^2\left(x\right)-{\sin }^2\left(x\right)}{\sin \left(x\right).\cos \left(x\right)}$

$=\frac{2\cos \left(2x\right)}{2\sin \left(x\right).\cos \left(x\right)}$ ($\mathit{c}\mathit{o}\mathit{s}\left(2x\right)\mathbf{=}\mathit{c}\mathit{o}{\mathit{s}}^{\mathbf{2}}\left(x\right)\mathbf{-}\mathit{s}\mathit{i}{\mathit{n}}^{\mathbf{2}}\left(x\right)$)

$=\frac{2\cos \left(2x\right)}{2\sin \left(x\right).\cos \left(x\right)}$

$=\frac{2\cos \left(2x\right)}{\sin \left(2x\right)}$ ($\mathit{s}\mathit{i}\mathit{n}\left(2x\right)\mathbf{=}\mathbf{2}\mathit{s}\mathit{i}\mathit{n}\left(x\right)\mathbf{.}\mathit{c}\mathit{o}\mathit{s}\left(x\right)$)

$\mathbf{=}\mathbf{2}\mathit{c}\mathit{o}\mathit{t}\left(2x\right)$

Hence, option (C) is correct.