Question

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

Answer 1

Solution :-

On taking LHS :

( 1-cos² A ) sec² B + tan² B ( 1-sin² A )

We know that

sin² A + cos² A = 1

sec² A - tan² A = 1

Now,

( 1- cos² A) sec² B + tan² B ( 1-sin² A ) becomes

=( sin² A ) (1 + tan² B ) + tan² B ( 1-sin² A )

=sin² A+sin² Atan² B+tan² B-tan² B sin² A

=sin² A+sin² Atan² B+tan² B-sin² Atan² B

=sin² A+tan² B+(sin² A tan² B-sin² Atan² B)

=sin² A + tan² B + (0)

=sin² A + tan² B

=RHS

Therefore, LHS = RHS

Hence, Proved.

Answer :-

( 1 - cos² A ) sec² B + tan² B ( 1 - sin² A )

= sin² A + tan² B

Used formulae:-

• sin² A + cos² A = 1

• sec² A - tan² A = 1

Answer 2

$\bold{ANSWER≈}$

On taking LHS:

(1-cos² A) sec² B + tan² B (1-sin² A)

We know that

sin² A + cos² A = 1

sec² A- tan² A = 1

Now,

(1- cos² A) sec² B + tan² B (1-sin² A) becomes

=( sin² A) (1 + tan² B) + tan² B (1-sin² A)

=sin² A+sin² Atan² B+tan² B-tan² B sin² A

=sin² A+sin² Atan² B+tan² B-sin² Atan² B

=sin² A+tan² B+(sin² A tan² B-sin² Atan² B)

=sin² A + tan² B + (0)

=sin² A + tan² B

=RHS

Therefore, LHS = RHS

Hence, Proved.

=sin² A+sin² Atan² B+tan² B-tan² B sin² A

=sin² A+sin² Atan² B+tan² B-sin² Atan² B

=sin² A+tan² B+(sin² A tan² B-sin² Atan² B)

=sin² A + tan² B + (0)

=sin² A + tan² B

=RHS

Therefore, LHS = RHS

Hence, Proved.

Answer :

(1-cos² A) sec² B + tan² B (1-sin² A)

= sin² A + tan² B

Used formulae:

sin² A+ cos² A = 1

• sec² A - tan² A = 1