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प्रश्न
Prove the identity (sin θ + cos θ)(tan θ + cot θ) = sec θ + cosec θ.
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उत्तर
L.H.S. = (sin θ + cos θ)(tan θ + cot θ)
= `(sin theta + cos theta)(sin theta/cos theta + costheta/sin theta)`
= `(sin theta + cos theta)((sin^2 theta + cos^2 theta)/(costhetasin theta))`
= `(sintheta+costheta)xx1/(sinthetacostheta)` ...[∵ sin2θ + cos2θ = 1]
= `(sin theta + cos theta)/(cos theta sin theta)`
= `sin theta/(cos thetasin theta) + cos theta/(cos theta sin theta)`
= `1/cos theta + 1/sin theta`
= `sec theta + cosec theta`
= R.H.S
Hence proved.
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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
