Question

prove that 1-tan² x/2 / 1+tan² x/2 = cosx.

Wrong and illogical answers will be reported immediately.....​

Answer 1

Answer:

proved

Step-by-step explanation:

LHS =  1-tan² x/2 / 1+tan² x/2 =  1-tan² x/2÷ 1+tan² x/2  

= 1 - sin²x/2/cos²x/2 ÷  1+sin²x/2/cos²x/2

= cos²x/2 - sin²x/2 ÷ cos²x/2 + sin²x/2

= cos²x/2 - sin²x/2

= cosx

= RHS

Answer 2

Answer:

$\red{ \frac{ 1-tan^{2} \frac{x}{2}}{1+tan^{2} \frac{x}{2}}}= \green { cos x}$

Step-by-step explanation:

$LHS =\red{ \frac{ 1-tan^{2} \frac{x}{2}}{1+tan^{2} \frac{x}{2}}}$

$Let \: A = \blue {\frac{x}{2}}\: ---(1)$

$= \frac{ 1-tan^{2} A}{1+tan^{2}A} \\= \frac{ 1-tan^{2}A}{sec^{2}A}$

$\boxed { \pink { 1+tan^{2}A = sec^{2}A}}$

$= \frac{ 1 - \frac{sin^{2}A}{cos^{2}A}}{\frac{1}{cos^{2}A}}$

$= cos^{2}A \left( 1- \frac{sin^{2}{A}}{cos^{2}A}\right)\\= cos^{2}A - sin^{2}A$

$= cos2A$

$\boxed { \pink { cos^{2}A - sin^{2}A = cos2A }}$

$= cos 2\times \frac{x}{2} \: \: [ from \:(1) ]$

$=\green { cos x} \\= RHS$

Therefore.,

$\red{ \frac{ 1-tan^{2} \frac{x}{2}}{1+tan^{2} \frac{x}{2}}}= \green { cos x}$

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