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प्रश्न
`(cos ec^theta + cot theta )/( cos ec theta - cot theta ) = (cosec theta + cot theta )^2 = 1+2 cot^2 theta + 2cosec theta cot theta`
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उत्तर
Here, `( cosec theta + cot theta )/( cosec theta - cot theta)`
= `((cosec theta + cot theta) ( cosec theta + cot theta ))/(( cosec theta - cot theta ) ( cosec theta + cot theta))`
=` ((cosec theta + cot theta)^2)/(( cosec ^2 theta - cot^2 theta))`
=`((cosec theta + cot theta )^2) /1`
=`(cosec theta + cot theta )^2`
Again , `( cosec theta + cot theta )^2`
= ` cosec^2 theta + cot^2 theta + 2 cosec theta cot theta `
=` 1+cot^2 theta + cot^2 theta + 2 cosec theta cot theta (∵ cosec^2 theta - cot^2 theta =1)`
=` 1+2 cot^2 theta + 2 cosec theta cot theta `
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संबंधित प्रश्न
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`tan^2A - tan^2B = (sin^2A - sin^2B)/(cos^2Acos^2B)`
Prove the following identity :
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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
