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प्रश्न
Prove the following trigonometric identities.
`(tan^3 theta)/(1 + tan^2 theta) + (cot^3 theta)/(1 + cot^2 theta) = sec theta cosec theta - 2 sin theta cos theta`
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उत्तर
`(tan^3 theta)/(1 + tan^2 theta) + (cot^3 theta)/(1 + cot^2 theta) ` [`∵ sec^2 theta - tan^2 theta = 1 - cosec^2 theta - cot^2 theta = 1`]
`= tan theta + cos^2 theta = cot^3 theta xx sin^3 theta`
`[∵ 1/sec^2 theta = cos^2 theta, 1/cosec^2 theta = 1 + cot^2 theta]`
`sin^3 theta/cos^3 theta xx cos^2 theta + cos^3 theta/sin^3 theta xx sin^2 theta`
`sin^3 theta/cos theta + cos^3 theta/sin theta`
`= (sin^4 theta + cos^4 theta)/(sin theta cos theta)`
` (1 - 2sin^2 theta cos^2 theta)/(sin theta cos theta)`
`1/(sin theta cos theta) - (2 sin^2 theta cos^2 theta)/(sin theta cos theta)`
`sec theta cosec theta - 2sin theta cos theta`.
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Show that, cotθ + tanθ = cosecθ × secθ
Solution :
L.H.S. = cotθ + tanθ
= `cosθ/sinθ + sinθ/cosθ`
= `(square + square)/(sinθ xx cosθ)`
= `1/(sinθ xx cosθ)` ............... `square`
= `1/sinθ xx 1/square`
= cosecθ × secθ
L.H.S. = R.H.S
∴ cotθ + tanθ = cosecθ × secθ
