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Question
Prove the following:
`tan^3x/(1 + tan^2x) + cot^3x/(1 + cot^2x)` = secx cosecx − 2sinx cosx
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Solution
L.H.S. = `tan^3x/(1 + tan^2x) + cot^3x/(1 + cot^2x)`
= `tan^3x/sec^2x + cot^3x/("cosec"^2x)`
= `sin^3x/cos^3x xx cos^2x + cos^3x/sin^3x xx sin^2x`
= `sin^3x/cosx + cos^3x/sinx`
= `(sin^4x + cos^4x)/(sinx*cosx)`
= `((sin^2x + cos^2x)^2 - 2sin^2x cos^2x)/(sinx* cosx)` ...[∵ a2 + b2 = (a + b)2 – 2ab]
= `(1 - 2sin^2x cos^2x)/(sinx*cosx)`
= `1/(sinx* cosx) - (2sin^2xcos^2x)/(sinx*cosx)`
= secx · cosecx – 2sinx · cosx
= R.H.S.
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