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प्रश्न
Prove that `(1 + sin "B")/"cos B" + "cos B"/(1 + sin "B")` = 2 sec B
उत्तर
L.H.S = `(1 + sin "B")/"cos B" + "cos B"/(1 + sin "B")`
= `((1 +sin "B")^2 + cos^2"B")/(cos "B"(1 + sin "B"))`
= `(1 +2sin"B" + sin^2"B" + cos^2"B")/(cos"B"(1 + sin"B"))` ......[∵ (a + b)2 = a2 + 2ab + b2]
= `(1 + 2sin"B" + 1)/(cos"B"(1+ sin"B"))` .....[∵ sin2B + cos2B = 1]
= `(2 + 2sin"B")/(cos"B"(1 + sin"B"))`
= `(2(1 + sin"B"))/(cos"B"(1 + sin"B"))`
= `2/"cos B"`
= 2 sec B
= R.H.S
∴ `(1 + sin "B")/"cos B" + "cos B"/(1 + sin "B")` = 2 sec B
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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
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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`
