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प्रश्न
Prove that `cot^2 "A" [(sec "A" - 1)/(1 + sin "A")] + sec^2 "A" [(sin"A" - 1)/(1 + sec"A")]` = 0
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उत्तर
L.H.S = `cot^2"A"[(sec"A" - 1)/(1 + sin "A")] + sec^2"A"[(sin"A" - 1)/(1 + sec"A")]`
= `(cot^2"A"(sec"A" - 1)(sec"A" + 1) + sec^2"A"(sin"A" - 1)(sin"A" + 1))/((1 + sin"A")(1 + sec"A"))`
= `(cot^2"A"(sec^2"A" - 1) + sec^2"A"(sin^2"A" - 1))/((1 + sin"A")(1 + sec"A"))`
= `(cot^2"A" xx tan^2"A" + sec^2"A"( - cos^2"A"))/((1 + sin"A")(1 + sec"A"))`
= `(cot^2"A" xx 1/cot^2"A" - sec^2"A" xx 1/sec^2"A")/((1 + sin"A")(1 + sec"A"))`
= `(1 - 1)/((1 + sin"A")(1+ sec"A"))`
= `0/((1 + sin"A")(1 + sec"A"))`
= 0
L.H.S = R.H.S
Hence it is proved.
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संबंधित प्रश्न
Prove the following trigonometric identities
`((1 + sin theta)^2 + (1 + sin theta)^2)/(2cos^2 theta) = (1 + sin^2 theta)/(1 - sin^2 theta)`
Prove the following trigonometric identities. `(1 - cos A)/(1 + cos A) = (cot A - cosec A)^2`
Prove the following identities:
`1/(tan A + cot A) = cos A sin A`
Prove the following identities:
`(costhetacottheta)/(1 + sintheta) = cosectheta - 1`
`sec theta (1- sin theta )( sec theta + tan theta )=1`
Prove the following identities:
`(sin theta + 1 - cos theta)/(cos theta - 1 + sin theta) = (1 + sin theta)/(cos theta)`
The value of \[\sqrt{\frac{1 + \cos \theta}{1 - \cos \theta}}\]
Prove the following identity :
`cosecA + cotA = 1/(cosecA - cotA)`
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`
Factorize: sin3θ + cos3θ
Hence, prove the following identity:
`(sin^3θ + cos^3θ)/(sin θ + cos θ) + sin θ cos θ = 1`
