Question

show that tan square teeta + tan square teeta = sec square teeta - sec square teeta​

Answer 1

$\bold\pink{Correct}$ $\bold\pink{Question:-}$

Show that

${ \tan }^{4} \theta \: + { \tan }^{2} \theta \: = { \sec }^{4} \theta \: - { \sec}^{2} \theta$

$\huge\bold{Solution:-}$

Given,

RHS =

${ \sec}^{4} \theta - { \sec}^{2} \theta$

${ \sec}^{2} \theta \: \: ( { \sec }^{2} \theta - 1)$

By formula:-

${ \sec}^{2} \theta \: = 1 + { \tan}^{2} \theta$

$1 + { \tan }^{2} \theta(1 + \tan^{2} \theta - 1)$

$1 + { \tan }^{2} \theta( { \tan}^{2} \theta)$

$= > { \tan}^{4} \theta + { \tan }^{2} \theta$

So,

LHS = RHS

Hence proved