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Answer 1

(i) (sin²A cos²B - cos²A sin²B) = (sin²A - sin²B)

Taking L.H.S.

=> sin²A (1 - sin²B) - (1 - sin²A) sin²B

=> sin²A - sin²B sin²A - (sin²B - sin²A sin²B)

=> sin²A - sin²B sin²A - sin²B + sin²A sin²B

=> sin²A - sin²B

L.H.S. = R.H.S.

________ [HENCE PROVED]

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(ii) (tan²A sec²B - sec²A tan²B) = (tan²A - tan²B)

Takin L.H.S.

=> tan²A (1 + tan²B) - (1 + tan²A) tan²B

=> tan²A + tan²A tan²B - (tan²B + tan²A tan²B)

=> tan²A + tan²A tan²B - tan²B - tan²A tan²B

=> tan²A - tan²B

L.H.S. = R.H.S.

________ [HENCE PROVED]

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✡ Formula used :

• cos² Ø = 1 - sin² Ø

• sec² Ø = 1 + tan² Ø

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Answer 2

$\large \:( \:i \:) \:( \:sin^{2} \:cos^{2}B \:- \:cos^{2}A \:sin^{2}B \:) \:= \:( \:sin^{2}A \:- \:sin^{2}B \:)$

$\large \:Taking \:L . H . S .$

$\large \:>> \:sin^{2}A \:( \:1 \:- \:sin^{2}B \:) \:- \:( \:1 \:- \:sin^{2}A \:) \:sin^{2}B$

$\large \:>> \:sin^{2}A \:- \:sin^{2}B \:sin^{2}A \:- \:( \:sin^{2}B \:- \:sin^{2}A \:sin^{2}B \:)$

$\large \:>> \:sin^{2}A \:- \:sin^{2}B \:sin^{2}A \:- \:sin^{2}B \:+ \:sin^{2}A \:sin^{2}B$

$\large \:sin^{2}A \:- \:sin^{2}B$

$\large \:L . H . S. \:= \:R . H . S .$

________________[Hence Proved]

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