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प्रश्न
Prove the following identity :
`1/(cosA + sinA - 1) + 2/(cosA + sinA + 1) = cosecA + secA`
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उत्तर
LHS = `1/((cosA + sinA) - 1) + 1/((cosA + sinA) + 1)`
= `(cosA + sinA + 1 + cosA + sinA - 1)/((cosA + sinA)^2 -1)`
= `(2(cosA + sinA))/(cos^2A + sin^2A + 2cosAsinA - 1)`
= `(2(cosA + sinA))/(1 + 2cosAsinA - 1) = (cosA + sinA)/(cosAsinA)`
= `cosA/(cosAsinA) + sinA/(cosAsinA)`
= `1/sinA + 1/cosA`
= cosecA + secA
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If sin θ + cos θ = `sqrt(3)`, then show that tan θ + cot θ = 1
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
