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प्रश्न
\[\frac{\tan \theta}{\sec \theta - 1} + \frac{\tan \theta}{\sec \theta + 1}\] is equal to
पर्याय
2 tan θ
2 sec θ
2 cosec θ
2 tan θ sec θ
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उत्तर
The given expression is `tan θ /(secθ-1)+tan θ/(sec θ+1)`
=` (tan θ (sec θ+1)+tan θ(secθ-1))/((secθ-1)(secθ+1))`
= `(tan θ sec θ+tanθ+tan θ secθ-tan θ)/(sec^2θ-1)`
=`( 2tanθ secθ)/tan^2θ`
=`(2secθ)/tan θ`
= `(2 1/cos θ)/(sinθ/cos θ)`
=`2 1/ sinθ`
= `2 cosec θ`
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संबंधित प्रश्न
(secA + tanA) (1 − sinA) = ______.
Prove the following trigonometric identities.
(1 + cot A − cosec A) (1 + tan A + sec A) = 2
Prove the following trigonometric identities.
`(cos A cosec A - sin A sec A)/(cos A + sin A) = cosec A - sec A`
If 3 sin θ + 5 cos θ = 5, prove that 5 sin θ – 3 cos θ = ± 3.
Prove that
`sqrt((1 + sin θ)/(1 - sin θ)) + sqrt((1 - sin θ)/(1 + sin θ)) = 2 sec θ`
If x=a `cos^3 theta and y = b sin ^3 theta ," prove that " (x/a)^(2/3) + ( y/b)^(2/3) = 1.`
Write the value of `3 cot^2 theta - 3 cosec^2 theta.`
Write True' or False' and justify your answer the following :
The value of sin θ+cos θ is always greater than 1 .
Prove the following identity :
`(cotA + tanB)/(cotB + tanA) = cotAtanB`
Prove the following identity :
`sqrt(cosec^2q - 1) = "cosq cosecq"`
Prove the following identity :
`sqrt((1 + sinq)/(1 - sinq)) + sqrt((1- sinq)/(1 + sinq))` = 2secq
Prove the following identity :
`1/(sinA + cosA) + 1/(sinA - cosA) = (2sinA)/(1 - 2cos^2A)`
Prove the following identity :
`tan^2θ/(tan^2θ - 1) + (cosec^2θ)/(sec^2θ - cosec^2θ) = 1/(sin^2θ - cos^2θ)`
Prove that `(sin θ tan θ)/(1 - cos θ) = 1 + sec θ.`
Prove the following identities.
`(sin^3"A" + cos^3"A")/(sin"A" + cos"A") + (sin^3"A" - cos^3"A")/(sin"A" - cos"A")` = 2
If sec θ = `25/7`, find the value of tan θ.
Solution:
1 + tan2 θ = sec2 θ
∴ 1 + tan2 θ = `(25/7)^square`
∴ tan2 θ = `625/49 - square`
= `(625 - 49)/49`
= `square/49`
∴ tan θ = `square/7` ........(by taking square roots)
Choose the correct alternative:
sec2θ – tan2θ =?
Prove that `(tan(90 - theta) + cot(90 - theta))/("cosec" theta)` = sec θ
Show that tan 7° × tan 23° × tan 60° × tan 67° × tan 83° = `sqrt(3)`
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`
