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Question
Prove the following trigonometric identities.
`sin A/(sec A + tan A - 1) + cos A/(cosec A + cot A + 1) = 1`
Solution
We have to prove `sin A/(sec A + tan A - 1) + cos A/(cosec A + cot A + 1) = 1`
We know that, `sin^2 A + cos^2 A = 1`
So,
`sin A/(sec A + tab A - 1) + cos A/(cosec A + cot A -1)`
`= sin A/(1/cos A + sin A/cos A - 1) + cos A/(1/sin A + cos A/sin A - 1)`
`= sin A/((1 + sin A - cos A)/cos A) + cos A/((1 + cos A - sin A)/sin A)`
`= (sin A cos A)/(1 + sin A - cos A) + (sin A cos A)/(1 + cos A - sin A)`
`= (sin A cos A(1 + cos A - sin A) + sin A cos A((1 + sin A - cos A)))/((1 + sin A - cos A)(1 + cos A- sin A))`
`= (sin A cos A (1 + cos A - sin A + 1 + sin A - cos A))/({1 + (sin A - cos A)}{1 - (sin A - cos A)})`
`= (2 sin A cos A)/(1 - (sin A - cos A)^2)`
`= (2 sin A cos A)/(1-(sin^2 A - 2 sin A cos A + cos^2 A))`
`= (2 sin A cos A)/(1 - (1 - 2 sin A cos A))`
`= (2 sin A cos A)/(1 - 1 + 2 sin A cos A)`
`= (2 sin A cos A)/(2 sin A cos A)`
= 1
Hence proved.
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