Question

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

Answer 1

Answer:

${tan}^{2} \alpha + {tan}^{4} \alpha \\ = {tan}^{2} \alpha (1 + {tan}^{2} \alpha )\\ = {tan}^{2} \alpha \times {sec}^{2} \alpha \\ = ( {sec}^{2} \alpha - 1)( {sec}^{2} \alpha ) \\ = {sec}^{4} \alpha - {sec}^{2} \alpha$

__ __ __ __ __ __ __ __

Identity used :-

$1 + {tan}^{2} \alpha = {sec}^{2} \alpha$