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Question
Prove the following trigonometric identities.
`1/(sec A - 1) + 1/(sec A + 1) = 2 cosec A cot A`
Solution 1
We need to prove `1/(secA - 1) + 1/(sec A + 1) = 2 cosec A cot A`
Solving the L.H.S, we get
`1/(sec A - 1)+ 1/(sec A + 1) = (sec A + 1 + sec A - 1)/((sec A - 1)(sec A + 1))`
`= (2 sec A)/(sec^2 A - 1)`
Further using the property ` 1 + tan^2 theta = sec^2 theta` we get
So
`(2 sec A)/(sec^2 A - 1) = (2 sec A)/(tan^2 A)`
`= (2(1/cos A))/(sin^2 A/cos^2 A)`
`= 2 1/cos A xx cos^2 A/sin^2 A`
`= 2(cos A/sin A) xx 1/sin A`
= 2cosec A cot A
Solution 2
LHS = `1/(sec A - 1) + 1/(sec A + 1)`
= `(sec A + 1 + sec A - 1)/(sec^2 A - 1 )`
= `(2sec A)/(tan^2 A)`
= `2 . 1/(cos A) xx 1/((sin^2 A)/(cos^2 A))`
= `2. 1/(cos A) xx (cos^2 A)/(sin^2 A)`
= 2 cosec A. cot A
= RHS
Hence proved.
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