Question

Sinx. Tanx/2+2cosx=2/1-tan^2x/2​

Answer 1

$\displaystyle\huge\red{\underline{\underline{Solution}}}$

CORRECT QUESTION

TO PROVE :

$\displaystyle \sf{ \sin x \:. \tan \frac{x}{2} + 2 \cos x = \frac{2}{1 + { \tan}^{2} \frac{x}{2} } \: }$

FORMULA TO BE IMPLEMENTED

$\displaystyle \sf{ 1. \: \: \sin x = \frac{2 \: \tan \frac{x}{2} }{1 + { \tan}^{2} \frac{x}{2} }}$

$\displaystyle \sf{ 2. \: \: \cos x = \frac{ 1 - { \tan}^{2} \frac{x}{2}}{1 + { \tan}^{2} \frac{x}{2}} \: }$

PROOF

LHS

$= \displaystyle \sf{ \sin x \:. \tan \frac{x}{2} + 2 \cos x \: }$

$= \displaystyle \sf{ \frac{2 \: \tan \frac{x}{2} }{1 + { \tan}^{2} \frac{x}{2} } \: .\tan \frac{x}{2} + 2 \times \frac{ 1 - { \tan}^{2} \frac{x}{2}}{1 + { \tan}^{2} \frac{x}{2}} \: }$

$= \displaystyle \sf{ \frac{2 \: {\tan}^{2} \frac{x}{2} }{1 + { \tan}^{2} \frac{x}{2} } \: + \frac{ 2 - 2 { \tan}^{2} \frac{x}{2}}{1 + { \tan}^{2} \frac{x}{2}} \: }$

$= \displaystyle \sf{ \frac{2 \: {\tan}^{2} \frac{x}{2} + 2 - 2 { \tan}^{2} \frac{x}{2} \: }{1 + { \tan}^{2} \frac{x}{2} } \: \: }$

$= \displaystyle \sf{ \frac{2 }{1 + { \tan}^{2} \frac{x}{2} } \: \: }$

= RHS

Hence proved

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Answer 2

Answer:

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