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प्रश्न
Prove that: `sqrt((1 - cos θ)/(1 + cos θ)) = "cosec" θ - cot θ`.
Prove the following:
`sqrt((1 - cos θ)/(1 + cos θ)) = "cosec" θ - cot θ`
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उत्तर
LHS = `sqrt((1 - cos θ)/(1 + cos θ) xx (1 - cos θ)/(1 - cos θ))`
= `sqrt((1 - cos θ)^2/(1 - cos^2θ))`
= `(1 - cos θ)/(sqrt(1 - cos^2θ))`
= `(1 - cos θ)/(sqrt(sin^2θ))`
= `(1 - cos θ)/(sin θ)`
= `(1)/(sin θ) - (cos θ)/(sin θ)`
= cosec θ − cot θ
= RHS
Hence proved.
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संबंधित प्रश्न
Prove that `sqrt((1 + cos theta)/(1 - cos theta)) + sqrt((1 - cos theta)/(1 + cos theta)) = 2 cosec theta`
`sin^2 theta + 1/((1+tan^2 theta))=1`
`cos^2 theta + 1/((1+ cot^2 theta )) =1`
Write the value of \[\cot^2 \theta - \frac{1}{\sin^2 \theta}\]
Write True' or False' and justify your answer the following:
\[ \cos \theta = \frac{a^2 + b^2}{2ab}\]where a and b are two distinct numbers such that ab > 0.
Prove the following identity :
`cos^4A - sin^4A = 2cos^2A - 1`
If sec θ = `25/7`, then find the value of tan θ.
If x sin3 θ + y cos3 θ = sin θ cos θ and x sin θ = y cos θ, then prove that x2 + y2 = 1
Prove the following:
`tanA/(1 + sec A) - tanA/(1 - sec A)` = 2cosec A
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`
