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​

Learn more:

70. prove that(cos A cosec A - sin A sec A)/cos A + sin A= cosec A ...

https://brainly.in/question/11611272

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

https://brainly.in/question/38346263

Answer 2

Step-by-step explanation:

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

• 1 - cos²A = sin²A

sin²A sec²B + tan²B - tan²A sin²A = sin²A + tan²B

sin²A ( sec²B - tan²A ) + tan²B = sin²A + tan²B

• sec²B - tan²A = 1

sin²A + tan²B = sin²A + tan²B