Question

8. Prove that (1 + tanA - secA) x (1 + tanA + SecA) = 2 tanA​

Answer 1

Step-by-step explanation:

Using (a+b)(a-b) = a² - b²

LHS =

(1 + tanA - secA) x (1 + tanA + SecA)

= ( 1 + tanA)² - sec²A

= 1 + tan²A + 2tanA - sec²A

Also, sec²A = 1 + tan²A

= 1 + tan²A + 2tanA - 1 - tan²A

= 2tanA = RHS

Hence, proved