Question

Solve

❌No Spamming❌

❌No Copied Answers ❌​

Answer 1

hope this helps you.. :)

Answer 2

$\huge\underline\mathcal{Solution}$

LHS:

${ \sec }^{2} theta - ( \frac{ { \sin}^{2}theta - 2 { \sin}^{4}theta }{2 { \cos }^{4} theta - { \cos }^{2}theta } )$

$= \frac{1}{ { \cos }^{2} theta} - \frac{1}{ { \cos }^{2} theta}( \frac{ { \sin }^{2}theta - 2 { \sin }^{4}theta }{2 { \cos }^{2} theta - 1} )$

$= \frac{1}{ { \cos }^{2} theta}(1 - \frac{( { \sin}^{2} theta - 2 { \sin }^{4} theta)}{2 { \cos}^{2}theta - 1 } )$

$= \frac{1}{ { \cos }^{2} theta}( \frac{2 { \cos }^{2} theta - 1 - { \sin}^{2}theta + 2 { \sin}^{4}theta}{2 { \cos }^{2}theta - 1 } )$

$= \frac{1}{ { \cos }^{2} theta}( \frac{2 { \cos }^{2}theta - ( { \sin}^{2}theta + { \cos}^{2}theta) - { \sin}^{2}theta + 2 { \sin }^{4}theta }{2 { \cos}^{2}theta - 1 } )$

$= \frac{1}{ { \cos }^{2} theta}( \frac{ { \cos}^{2}theta - { \sin }^{2}theta - { \sin}^{2}theta + 2 { \sin}^{4} theta }{2 { \cos }^{2}theta - 1} )$

$= \frac{1}{ { \cos }^{2} theta}( \frac{ { \cos }^{2}theta - 2 { \sin}^{2}theta + 2 { \sin }^{4}theta }{2 { \cos}^{2}theta - 1 } )$

$= \frac{1}{ { \cos }^{2} theta}( \frac{ { \cos }^{2}theta - 2 { \sin }^{2}theta(1 - { \sin }^{2}theta) }{2 { \cos}^{2}theta - 1 } )$

$= \frac{1}{ { \cos }^{2} theta} \times \frac{( { \cos}^{2}theta - 2 { \sin }^{2}theta \: { \cos }^{2}theta)}{2 { \cos }^{2}theta - 1 }$

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

$= \frac{1 - 2(1 - { \cos}^{2}theta )}{2 { \cos }^{2}theta - 1 }$

$= \frac{1 - 2 + 2 { \cos}^{2}theta }{2 { \cos }^{2}theta - 1 }$

$= \frac { 2 { \cos }^{2}theta - 1}{2 { \cos}^{2}theta - 1 }$

$= 1$

= RHS

Hence proved