Question

(1 - cos²A) sec²B + tan²B (1-sin²A) = sin²A + tan²B​

Answer 1

Given : ( 1 − cos²A ) . sec²B + tan²B ( 1− sin²A ) = sin²A + tan²B​

To Find : Prove

Solution:

( 1 − cos²A ) . sec²B + tan²B ( 1− sin²A ) = sin²A + tan²B​

LHS =

( 1 − cos²A ) . sec²B + tan²B ( 1− sin²A )

using 1 − cos²A = sin²A

= sin²Asec²B + tan²B ( 1− sin²A )

Apply distributive property

= sin²Asec²B + tan²B   1− tan²Bsin²A

take  sin²A common in 1st and 3rd term

=  sin²A(sec²B  - tan²B)  +  tan²B

sec²B  - tan²B = 1

= sin²A(1)  +  tan²B

=  sin²A  +  tan²B

= RHS

QED

Hence proved

( 1 − cos²A ) . sec²B + tan²B ( 1− sin²A ) = sin²A + tan²B​

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Answer 2

Solution :-

→ (1 - cos²A) * sec²B + tan²B (1 - sin²A) = sin²A + tan²B

solving LHS,

→ (1 - cos²A) * sec²B + tan²B (1 - sin²A)

using :-

1 - cos²A = sin²A

→ sin²A * sec²B + tan²B (1 - sin²A)

→ sin²A * sec²B + tan²B - tan²B * sin²A

→ sin²A * sec²B - tan²B * sin²A + tan²B

→ sin²A(sec²B - tan²B) + tan²B

using :-

sec²B - tan²B = 1

→ sin²A * 1 + tan²B

→ sin²A + tan²B = RHS (Proved).

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