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प्रश्न
Prove that (1 – cos2A) . sec2B + tan2B(1 – sin2A) = sin2A + tan2B
उत्तर
L.H.S = (1 – cos2A) . sec2B + tan2B(1 – sin2A)
= `sin^2"A"* 1/(cos^2"B") + (sin^2"B")/(cos^2"B") (1 - sin^2"A")` ......`[(because sin^2"A" + cos^2"A" = 1),(therefore 1 - cos^2"A" = sin^2"A")]`
= `(sin^2"A")/(cos^2"B") + (sin^2"B")/(cos^2"B") - (sin^2"A"sin^2"B")/(cos^2"B")`
= `(sin^2"A")/(cos^2"B") - (sin^2"A"sin^2"B")/(cos^2"B") + (sin^2"B")/(cos^2"B")`
= `(sin^2"A")/(cos^2"B") (1 - sin^2"B") + tan^2"B"`
= `(sin^2"A")/(cos^2"B") (cos^2"B") + tan^2"B"`
= sin2A + tan2B
= R.H.S
∴ (1 – cos2A) . sec2B + tan2B (1 – sin2A) = sin2A + tan2B
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