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प्रश्न
Prove that `"cosec" θ xx sqrt(1 - cos^2θ) = 1`.
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उत्तर
L.H.S. = `"cosec" θ xx sqrt(1 - cos^2θ)`
= `"cosec" θ xx sqrt(sin^2θ)` ...`[(∵ sin^2θ + cos^2θ = 1),(therefore 1 - cos^2θ = sin^2θ)]`
= cosec θ × sin θ
= 1 ...[∵ sin θ × cosec θ = 1]
= R.H.S.
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संबंधित प्रश्न
Prove the following trigonometric identities.
`cos theta/(1 + sin theta) = (1 - sin theta)/cos theta`
Prove the following trigonometric identities.
`[tan θ + 1/cos θ]^2 + [tan θ - 1/cos θ]^2 = 2((1 + sin^2 θ)/(1 - sin^2 θ))`
Prove the following trigonometric identities.
tan2 A sec2 B − sec2 A tan2 B = tan2 A − tan2 B
Prove the following identities:
`(1 + sin A)/(1 - sin A) = (cosec A + 1)/(cosec A - 1)`
Prove the following identities:
sec4 A (1 – sin4 A) – 2 tan2 A = 1
`(1+ cos theta)(1- costheta )(1+cos^2 theta)=1`
`cot^2 theta - 1/(sin^2 theta ) = -1`a
`1+((tan^2 theta) cot theta)/(cosec^2 theta) = tan theta`
cosec4 θ − cosec2 θ = cot4 θ + cot2 θ
If `cos theta = 2/3 , "write the value of" ((sec theta -1))/((sec theta +1))`
Write True' or False' and justify your answer the following :
The value of \[\sin \theta\] is \[x + \frac{1}{x}\] where 'x' is a positive real number .
\[\frac{1 - \sin \theta}{\cos \theta}\] is equal to
\[\frac{\tan \theta}{\sec \theta - 1} + \frac{\tan \theta}{\sec \theta + 1}\] is equal to
Prove the following identity :
`cos^4A - sin^4A = 2cos^2A - 1`
Prove the following identity :
`(cosecθ)/(tanθ + cotθ) = cosθ`
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`tan ((B + C)/2) = cot "A/2`
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1 + cot2θ = ?
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`
