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प्रश्न
Prove the following identities:
`cot^2A((secA - 1)/(1 + sinA)) + sec^2A((sinA - 1)/(1 + secA)) = 0`
उत्तर
`cot^2A((secA - 1)/(1 + sinA)) + sec^2A((sinA - 1)/(1 + secA))`
= `cot^2A((secA - 1)/(1 + sinA) xx (secA + 1)/(secA + 1)) + sec^2A((sinA - 1)/(1 + secA))`
= `cot^2A[(sec^2A - 1)/((1 + sinA)(secA + 1))] + sec^2A((sinA - 1)/(1 + secA))`
= `cot^2A[(tan^2A)/((1 + sinA)(secA + 1))] + sec^2A((sinA - 1)/(1 + secA))`
= `1/((1 + sinA)(secA + 1)) + sec^2A((sinA - 1)/(1 + secA))`
= `(1 + sec^2A(sinA - 1)(1 + sinA))/((1 + sinA)(secA + 1))`
= `(1 + sec^2A(sin^2A - 1))/((1 + sinA)(secA + 1))`
= `(1 + sec^2A(-cos^2A))/((1 + sinA)(secA + 1))`
= `(1 - 1)/((1 + sinA)(secA + 1))`
= 0
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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)`
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∴ sin θ = `9/41`
The value is cosec θ = `41/9`, and sin θ = `9/41`
Show that, cotθ + tanθ = cosecθ × secθ
Solution :
L.H.S. = cotθ + tanθ
= `cosθ/sinθ + sinθ/cosθ`
= `(square + square)/(sinθ xx cosθ)`
= `1/(sinθ xx cosθ)` ............... `square`
= `1/sinθ xx 1/square`
= cosecθ × secθ
L.H.S. = R.H.S
∴ cotθ + tanθ = cosecθ × secθ
