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प्रश्न
If cosec θ + cot θ = p, then prove that cos θ = `(p^2 - 1)/(p^2 + 1)`
उत्तर
According to the question,
cosec θ + cot θ = p
Since, cosec θ = `1/sintheta` and cot θ = `costheta/sintheta`
`1/sintheta + costheta/sintheta` = p
`(1 + costheta)/sintheta` = p
Squaring on L.H.S and R.H.S,
`((1 + costheta)/sin theta)^2` = p2
`(1 + cos^2 theta + 2 cos theta)/(sin^2 theta)` = p2
Applying component and dividend rule,
`((1 + cos^2 theta + 2 cos theta) - sin^2 theta)/((1 + cos^2 theta + 2 cos theta) + sin^2 theta) = ("p"^2 - 1)/("p"^2 + 1)`
= `((1 - sin^2theta) + cos^2 theta + 2 cos theta)/(sin^2 theta + cos^2 theta + 1 + 2 cos theta) = ("p"^2 - 1)/("p"^2 + 1)`
Since, 1 – sin2θ = cos2θ and sin2θ + cos2θ = 1
`(cos^2 theta + cos^2 theta + 2 cos theta)/(1 + 1 + 2 cos theta) = ("p"^2 - 1)/("p"^2 + 1)`
`(2 cos^2 theta + 2 cos theta)/(2 + 2 cos theta) = ("p"^2 - 1)/("p"^2 + 1)`
`(2 cos theta(cos theta + 1))/(2(cos theta + 1)) = ("p"^2 - 1)/("p"^2 + 1)`
cos θ = `("p"^2 - 1)/("p"^2 + 1)`
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
