Question

(sec A - cos A) (sec A + cos A)
= sinº A + tan? A​

Answer 1

Answer:

Step-by-step explanation:

Given,

LHS = (sec A - cos A) (sec A + cos A)

=( sec ^2 A - cos^2 A) [since (a-b) (a+b) = a^2 -b^2 ]

= ( 1 + tan ^2 A - cos ^2 A ) [∵sec^2 A = 1 + tan ^2 A ]

= (1 - cos ^2 A + tan ^2 A )

= sin^ 2 A + tan^2 A = RHS

∴LHS = RHS

Hence (sec A - cos A ) (sec A + cos A ) = sin^2 A + tan^2 A is proved.