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Question
Prove the following identity :
`sqrt((1 - cosA)/(1 + cosA)) = sinA/(1 + cosA)`
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Solution
= LHS = `sqrt((1 - cosA)/(1 - cosA))`
= `sqrt((1 - cosA)/(1 + cosA) . (1 + cosA)/(1 + cosA))`
= `sqrt((1 - cos^2A)/(1 + cosA)^2)`
= `sqrt(sin^2A/(1 + cosA)^2)`
= `sinA/(1 + cosA)`
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RELATED QUESTIONS
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`(1 + cos A)/sin^2 A = 1/(1 - cos A)`
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`sin A/(sec A + tan A - 1) + cos A/(cosec A + cot A + 1) = 1`
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(sin A + cosec A)2 + (cos A + sec A)2 = 7 + tan2 A + cot2 A
Prove the following identities:
`1 - cos^2A/(1 + sinA) = sinA`
`(1+ cos theta + sin theta)/( 1+ cos theta - sin theta )= (1+ sin theta )/(cos theta)`
Prove that `((1 + sin θ - cos θ)/( 1 + sin θ + cos θ))^2 = (1 - cos θ)/(1 + cos θ)`.
Prove that `(tan θ + sin θ)/(tan θ - sin θ) = (sec θ + 1)/(sec θ - 1)`
If `cos theta/(1 + sin theta) = 1/"a"`, then prove that `("a"^2 - 1)/("a"^2 + 1)` = sin θ
The value of 2sinθ can be `a + 1/a`, where a is a positive number, and a ≠ 1.
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
