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प्रश्न
Prove the following identity :
`(1 + cosA)/(1 - cosA) = tan^2A/(secA - 1)^2`
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उत्तर
LHS = `(1 + cosA)/(1 - cosA)`
= `(1 + 1/secA)/(1 - 1/secA) = (secA + 1)/(secA - 1)`
= `(secA + 1)/(secA - 1) . (secA - 1)/(secA - 1)`
= `(sec^2A - 1)/(secA - 1)^2 = tan^2A/(secA - 1)^2` (`Q sec^2A - 1 = tan^2A`)
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संबंधित प्रश्न
if `cos theta = 5/13` where `theta` is an acute angle. Find the value of `sin theta`
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`cos theta/(1 - sin theta) = (1 + sin theta)/cos theta`
Prove the following trigonometric identities.
sin2 A cot2 A + cos2 A tan2 A = 1
Prove the following identities:
`1/(tan A + cot A) = cos A sin A`
Prove the following identities:
`(sinA + cosA)/(sinA - cosA) + (sinA - cosA)/(sinA + cosA) = 2/(2sin^2A - 1)`
Prove the following identities:
`(sec"A"-1)/(sec"A"+1)=(sin"A"/(1+cos"A"))^2`
prove that `1/(1 + cos(90^circ - A)) + 1/(1 - cos(90^circ - A)) = 2cosec^2(90^circ - A)`
sin4A – cos4A = 1 – 2cos2A. For proof of this complete the activity given below.
Activity:
L.H.S. = `square`
= (sin2A + cos2A) `(square)`
= `1 (square)` ...`[sin^2"A" + square = 1]`
= `square` – cos2A ...[sin2A = 1 – cos2A]
= `square`
= R.H.S.
If cos A = `(2sqrt(m))/(m + 1)`, then prove that cosec A = `(m + 1)/(m - 1)`.
If cot θ = `40/9`, find the values of cosec θ and sinθ,
We have, 1 + cot2θ = cosec2θ
1 + `square` = cosec2θ
1 + `square` = cosec2θ
`(square + square)/square` = cosec2θ
`square/square` = cosec2θ ......[Taking root on the both side]
cosec θ = `41/9`
and sin θ = `1/("cosec" θ)`
sin θ = `1/square`
∴ sin θ = `9/41`
The value is cosec θ = `41/9`, and sin θ = `9/41`
