Question

plzzz solve this fast

Answer 1

Hello Friend,

Here, we will use the identity:

sec²θ = 1 + tan²θ

Here,
LHS
= tan²A sec²B - sec²A tan²B
= tan²A (1 + tan²B) - tan²B (1 + tan²A)
= tan²A + tan²A tan²B - tan²B - tan²A tan²B
= tan² A - tan² B + (tan²A tan²B - tan²A tan²B)
= tan² A - tan² B + 0
= tan²A - tan²B
= RHS

Hence proved.

Hope it helps.

Purva
@Purvaparmar1405
Brainly.in

Answer 2

tan²α×sec²β-sec²α×tan²β=tan²α-tan²β                                (A=α & B=β)
Using,
sec²θ = 1 + tan²θ

solving L.H.S, we get
= tan²α sec²β - sec²α tan²β
= tan²α(1+tan²β)-tan²β(1+tan²α)
= tan²α+tan²α tan²β-tan²β-tan²α tan²β
= tan²α-tan²β+(tan²α tan²β-tan²α tan²β)
= tan²α-tan²β + 0
= tan²α-tan²β
= R.H.S.
Hence proved