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प्रश्न
Prove the following trigonometric identities. `(1 - cos A)/(1 + cos A) = (cot A - cosec A)^2`
उत्तर १
We need to prove `(1 - cos A)/(1 + cos A) = (cot A - cosec A)^2`
Now, rationalising the L.H.S, we get
`(1 - cos A)/(1 + cos A) = ((1 - cos A)/(1 + cos A)) ((1 - cos A)/(1 - cos A))`
`= (1 - cos A)^2/(1 - cos^2 A)` (using `a^2 - b^2 = (a + b)(a - b))`
` = (1 + cos^2 A - 2 cos A)/sin^2 A` (Using `sin^2 theta = 1 - cos^2 theta`)
`= 1/sin^2 A + cos^2 A/sin^2 A - (2 cos A)/sin^2 A`
Using `cosec theta = 1/sin theta` and `cot theta = cos theta/sin theta` we get
`1/sin^2 A + cos^2 A/sin^2 A - (2 cos A)/sin^2 A = cosec^2 A + cot^2 A - 2 cot A cosec A`
` (cot A - cosec A)^2` (Using `(a + b)^2 = a^2 + b^2 + 2ab`)
Hence proved.
उत्तर २
LHS = `(1 - cos θ)/(1 + cos θ)`
= `(1 - cos θ)/(1 + cos θ) xx (1 - cos θ)/(1 - cos θ)`
= `(1 - cos θ)^2/(1 - cos^2 θ)`
= `(1 - cos θ)^2/(sin^2 θ)`
= `[(1 - cosθ)/(sin θ)]^2`
= `[ 1/sinθ - cosθ/sin θ ]^2`
= ( cosec θ - cot θ )2
= [ - (cot θ - cosec θ)]2
= (cot θ - cosec θ)2
= RHS
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
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