Question

7. Prove that
(1 - cos2A) × sec2B + tan2’B (1-sin2A) = sin2A + tan2B.
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Answer 1

Step-by-step explanation:

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Answer 2

Answer:

Prove that

(1.) ( 1-cos²A). sec²B + tan² B (1-sin²A) = sin² A + tan² B

Answer :- ⤵️

Given , cos²A - sin²A = tan²B

→ cos² A - (1-cos²A) = tan²B [ sin² ¢ = 1- cos²¢]

→ cos² A - 1 + cos² A = tan²B

→ 2cos² A - 1 = tan²B

→ 2cos² A = 1 + tan² B

→ 2 cos² A = sec² B [ sec² ¢ = 1+ tan² ¢]

→ 2 cos² A. cos² B = 1 [ sec ¢ =1/cos¢]

→ 2 cos² B = sec² A

→ cos² B + cos² B = 1 + tan² A

→ cos² B + cos² B -1 = tan² A

→ cos² B -(1-cos² B) = tan²A

→ cos²B - sin² B = tan² A

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