Question

Prove:cot2x-tanx=cos3x/cosx×sin2x​

Answer 1

$\textbf{\underline{\underline{According\:to\:the\:Question}}}$

LHS - cot2x - tanx

$\tt{\rightarrow\dfrac{1}{tan2x} - tanx}$

$\tt{\rightarrow\dfrac{1-tanxtan2x}{tan2x}}$

$\tt{\rightarrow\dfrac{1-sinxsin2x/cosxcos2x}{tan2x}}$

$\tt{\rightarrow\dfrac{cosxcos2x-sinxsin2x}{cosxcos2x\times sin2x/cos2x}}$

$\tt{\rightarrow\dfrac{cos(x+2x)}{cosxsin2x}}$

$\tt{\rightarrow\dfrac{cos3x}{cosxsin2x}}$

LHS = RHS

Answer 2

Answer: The following equation  can be shown true

Step-by-step explanation:

Given,

$\cot2x-\tan x\\\;\\=\frac{\cos2x}{\sin2x}-\frac{\sin x}{\cos x}\\\;\\=\frac{\cos2x.\cos x-\sin x.\sin2x}{\sin 2x.\cos x}\\\;\\=\frac{\cos(x+2x)}{\sin2x.\cos x}\;\;\;\;[\because\;\cos(A+B)=\cos A.\cos B-\sin A\sin B]\\\;\\=\frac{\cos3x}{\sin2x\cos x}\\\;\\\cot2x-\tan x=\frac{\cos3x}{\sin2x\cos x}$

Hence Proved.

Note:-

$1.) \cot x=\frac{\cos x}{\sin x}\\\;\\2.)\tan x=\frac{\sin x}{\cos x}\\\;\\3.)\cos(A+B)=\cos A.\cos B-\sin A\sin B\\\;\\4.)\cos(A-B)=\cos A.\cos B+\sin A\sin B$