Question

Prove that : sec^4 theta -sec^2 theta = tan^4 theta + tan ^2 theta

Answer 1

Solution:

Given LHS = $sec^{4}\theta -sec^{2}\theta$

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

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

/* By Trigonometric identity:

sec²A = 1+tan²A */

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

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

Rearranging the terms, we get

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

= RHS

••••

Answer 2

Answer:

Sry for the handwriting ☺️

Step-by-step explanation:

Id used :-

Sec^2 A = tan^2 A + 1