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प्रश्न
If A + B = 90°, prove that `(tan"A" tan"B" + tan"A" cot"B")/(sin"A" sec"B") - (sin^2"B")/(cos^2"A")` = tan2A
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उत्तर
A + B = 90°
⇒ B = 90° - A
L.H.S.
= `(tan"A" tan"B" + tan"A" cot"B")/(sin"A" sec"B") - (sin^2"B")/(cos^2"A")`
= `(tan"A" tan(90° - "A") + tan"A" cot(90° - "A"))/(sin"A" "cosec""A") - (sin^2 (90° - "A"))/(cos^2 "A")`
= `(tan"A" cot"A" + tan"A" tan"A")/(sin"A" "cosec""A") - (cos^2"A")/(cos^2"A")`
= `( 1 + tan^2 "A")/(1) - 1`
= 1 + tan2A - 1
= tan2A
= R.H.S.
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