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प्रश्न
Prove the following trigonometric identities.
`1/(sec A - 1) + 1/(sec A + 1) = 2 cosec A cot A`
उत्तर १
We need to prove `1/(secA - 1) + 1/(sec A + 1) = 2 cosec A cot A`
Solving the L.H.S, we get
`1/(sec A - 1)+ 1/(sec A + 1) = (sec A + 1 + sec A - 1)/((sec A - 1)(sec A + 1))`
`= (2 sec A)/(sec^2 A - 1)`
Further using the property ` 1 + tan^2 theta = sec^2 theta` we get
So
`(2 sec A)/(sec^2 A - 1) = (2 sec A)/(tan^2 A)`
`= (2(1/cos A))/(sin^2 A/cos^2 A)`
`= 2 1/cos A xx cos^2 A/sin^2 A`
`= 2(cos A/sin A) xx 1/sin A`
= 2cosec A cot A
उत्तर २
LHS = `1/(sec A - 1) + 1/(sec A + 1)`
= `(sec A + 1 + sec A - 1)/(sec^2 A - 1 )`
= `(2sec A)/(tan^2 A)`
= `2 . 1/(cos A) xx 1/((sin^2 A)/(cos^2 A))`
= `2. 1/(cos A) xx (cos^2 A)/(sin^2 A)`
= 2 cosec A. cot A
= RHS
Hence proved.
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If x = a sec θ + b tan θ and y = a tan θ + b sec θ prove that x2 - y2 = a2 - b2.
Prove that sin θ sin( 90° - θ) - cos θ cos( 90° - θ) = 0
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Prove the following:
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`1 + (cot^2 alpha)/(1 + "cosec" alpha)` = cosec α
If cosec θ + cot θ = p, then prove that cos θ = `(p^2 - 1)/(p^2 + 1)`
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`
