Question

please give me answer of maths question​ 5​

Answer 1

Question: If tan^2 alpha = 1 + 2 tan^2 beta. Prove that 2sin^2 alpha = 1 + sin^2 beta

Answer:

tan²A = 1 + 2tan²B

Adding 1 to both sides :

=> 1 + tan²A = 1 + 1 + 2 tan²B

=> 1 + tan²A = 2(1 + tan²B)

=> sec²A = 2sec²B

=> (1/cos²A) = 2(1/cos²B)

=> cos²B = 2cos²A

=> 1 - sin²B = 2(1 - sin²A)

=> 1 - sin²B = 2 - 2sin²A

=> 2sin²A = 2 - 1 + sin²B

=> 2sin²A = 1 + sin²B

Proved using:

• 1 + tan²x = sec²x

• secx = 1/cosx => sec²x = 1/cos²x

• cos²x = 1 - sin²x

*alpha is written as A and beta as B.

Answer 2

$\sf \: { \red{to \: prove}} \\ \sf \: tan² \: A = 1 + 2tan² \: B \\ \\ \sf \: { \red{proof}} \\ \\ \sf \: Adding \: 1 \: to \: both \: sides : \\ \sf \: 1 + tan²A = 1 + 2 tan²B \\ \\ \sf \: 1 + tan²A = 2(1 + tan²B) \\ \\ \sf \: sec²A = 2sec²B \\ \\ \sf \ ( \frac{1}{cos²A} ) = 2( \frac{1}{cos²B} ) \\ \\ \sf \: cos²B = 2cos²A \\ \\ \sf \: 1 - sin²B = 2(1 - sin²A) \\ \\ \sf \: 1 - sin²B = 2 - 2sin²A \\ \\ \sf \: 2sin²A = 2 - 1 + sin²B \\ \\ \sf \: 2sin²A = 1 + sin²B$

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