Question

Prove tan ^2 x +tan^4 x= sec^4 x -sec^2 x

Answer 1

${\underline{{\text{To Prove:}}}}$

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

${\underline{{\text{Solution:}}}}$

${ \tan }^{4} x + { \tan }^{2} x \\ \\ = { \tan}^{2} ({ \tan }^{2} x + 1) \\ \\ = ( { \sec}^{2} x - 1)({ \sec}^{2} x) \\ \\ = \sec {}^{4}x - { \sec}^{2} x \\ \\ \therefore \text{L.H.S = R.H.S [Proved]}$

${{\boxed{ \text{\blue{Some important trigonometry Rules}}}}}$

$cosec\ \theta\ . \sin\ \theta\ =\ 1\\\\tan\ \theta\ .\ cot\ \theta\ =\ 1\\\\sin^2\ \theta\ +\ cos^2\ theta\ =\ 1\\\\cosec^2\ \theta\ -\ cot^2\ theta\ =\ 1\\\\sec^2\ \theta\ -\ tan^2\ \theta\ =\ 1$