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प्रश्न
Prove the following identities:
`sqrt((1 - cosA)/(1 + cosA)) = sinA/(1 + cosA)`
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उत्तर
`sqrt((1 - cosA)/(1 + cosA))`
= `sqrt((1 - cosA)/(1 + cosA) xx (1 + cosA)/(1 + cosA))`
= `sqrt((1 - cos^2A)/(1 + cosA)^2)`
= `sqrt(sin^2A/(1 + cosA)^2)`
= `sinA/(1 + cosA)`
संबंधित प्रश्न
Prove that: `(1 – sinθ + cosθ)^2 = 2(1 + cosθ)(1 – sinθ)`
`(1+ cos theta + sin theta)/( 1+ cos theta - sin theta )= (1+ sin theta )/(cos theta)`
Write the value of `(cot^2 theta - 1/(sin^2 theta))`.
Write the value of ` sec^2 theta ( 1+ sintheta )(1- sintheta).`
Prove the following identity :
`(sinA + cosA)/(sinA - cosA) + (sinA - cosA)/(sinA + cosA) = 2/(2sin^2A - 1)`
Without using trigonometric table , evaluate :
`sin72^circ/cos18^circ - sec32^circ/(cosec58^circ)`
Prove that `(sin^2theta)/(cos theta) + cos theta` = sec θ
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
To prove cot θ + tan θ = cosec θ × sec θ, complete the activity given below.
Activity:
L.H.S = `square`
= `square/sintheta + sintheta/costheta`
= `(cos^2theta + sin^2theta)/square`
= `1/(sintheta*costheta)` ......`[cos^2theta + sin^2theta = square]`
= `1/sintheta xx 1/square`
= `square`
= R.H.S
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`
