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Maharashtra State BoardSSC (English Medium) 10th Standard

Prove that (1 – cos^2A) . sec^2B + tan^2B (1 – sin^2A) = sin^2A + tan^2B.

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Question

Prove that (1 – cos2A) . sec2B + tan2B (1 – sin2A) = sin2A + tan2B.

Theorem
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Solution

L.H.S. = (1 – cos2A) . sec2B + tan2B (1 – sin2A)

= `sin^2A * 1/(cos^2B) + (sin^2B)/(cos^2B) (1 - sin^2A)`   ...`[(∵ sin^2A + cos^2A = 1),(∴ 1 - cos^2A = sin^2A)]`

= `(sin^2A)/(cos^2B) + (sin^2B)/(cos^2B) - (sin^2A sin^2B)/(cos^2B)`

= `(sin^2A)/(cos^2B) - (sin^2A sin^2B)/(cos^2B) + (sin^2B)/(cos^2B)`

= `(sin^2A)/(cos^2B) (1 - sin^2B) + tan^2B`

= `(sin^2A)/(cos^2B) (cos^2B) + tan^2B`

= sin2A + tan2B

= R.H.S.

∴ (1 – cos2A) . sec2B + tan2B (1 – sin2A) = sin2A + tan2B

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Chapter 6: Trigonometry - Exercise

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