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प्रश्न
Prove the following trigonometric identities.
`(cosec A)/(cosec A - 1) + (cosec A)/(cosec A = 1) = 2 sec^2 A`
उत्तर
We need to prove `(cosec A)/(cosec A - 1) + (cosec A)/(cosec A = 1) = 2 sec^2 A`
Using identity `a^2 - b^2 = (a + b)(a - b)` we get
`(cosec A)/(cosec A - 1) = (cosec A)/(cosec A + 1) = (cosec A(cosec A + 1)+cosec A(cosec A - 1))/(cosec^2 A - 1)`
`= (cosec A (cosec A +1 + cosec A - 1))/(cosec^2 A - 1)`
Further, using the property `1 + cot62 theta = cosec^2 theta` we get
So
`(cosec A (cosec A + 1 + cosec A - 1))/(cosec^2 A- 1) = (cosec A(2 cosec A))/cot^2 A`
`= (2cosec^2 A)/cot^2 A`
`= (2)(1/sin^2 A)((cos^2 A)/(sin^2 A))`
`= 2(1/cos^2 A)`
`= 2 sec^2 A`
Hence proved.
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