Question

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

Answer 1

हल -

बायाँ पक्ष :

$\tan {}^{4} \theta + tan {}^{2} \theta \: \\ \\ = {tan}^{2} \theta(tan {}^{2} \theta + 1) \\ \\ \\ = (sec {}^{2} \theta - 1)sec {}^{2} \theta \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: (1 + {tan}^{2} \theta = {sec}^{2} \theta \\ \\ \\ \\ {sec}^{4} \theta- {sec}^{2} \theta \:$


इति सिध्दम

Answer 2

hope this will help uh ....
nd plz mark as brainliest ....