Question

Prove that ( tanA - tanB )^2 + ( 1 + tanA.tanB)^2 = sec^2A. sec^2B.​

Answer 1

Answer:

Step-by-step explanation:

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Answer 2

Answer:

(tanA - tanB)² + (1 + tanA.tanB)²

【 ( a - b )² = a² + b² - 2ab 】

【 ( a + b )² = a² + b² + 2ab 】

( tan²A + tan²B - 2 . tanA . tanB ) + (1 + tan²A . tan²B + 2 . tanA . tanB

Here ,

-2 . tanA . tanB & +2 . tanA . tanB gets cancelled

1 + tan²A + tan²B + tan²A . tan²B

( 1 + tan²A ) . ( 1 + tan²B )

sec²A . sec²B

Hence proved .