Question

cos^2theta - sin^2 theta = tan^2 alpha
then prove that
cos^2 alpha - sin^2 alpha = tan^2 theta
(sorry the previous question was rong so i have reposted the correct question)
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Answer 1

Given :-

• cos^2A - sin^2A = tan^2B

To Prove :-

• cos^2B - sin^2B = tan^2A

Solution :-

→ cos²A - sin²A = tan²B

Putting (sin²x = 1 - cos²x) in LHS we get,

→ cos²A - ( 1 - cos²A ) = tan²B

→ cos²A - 1 + cos²A = tan²B

→ 2cos²A - 1 = tan²B

→ 2cos²A = 1 + tan²B

Putting sec²x = (1 + tan²x) in RHS Now,

→ 2cos²A = sec²B

Putting sec x = (1/cosx) in RHS now,

→ 2cos²A = (1/cos²B)

→ 2cos²A * cos²B = 1

→ 2(1/sec²A) * cos²B = 1

→ 2cos²B = sec²A

→ cos²B + cos²B = 1 + tan²A

→ cos²B + cos²B - 1 = tan²A

Taking (-1) common in LHS Now,

→ cos²B - (1 - cos²B) = tan²A

Putting (1 - cos²x) = sin²x in Last ,

→ cos²B - sin²B = tan²A (Proved).

Answer 2

Answer:

cos 2b-sin 2b=tan 2a

Step-by-step explanation:

HOPE IT WILL HELP YOU