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Question
Prove the following trigonometric identities.
`(sec A - tan A)/(sec A + tan A) = (cos^2 A)/(1 + sin A)^2`
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Solution
We need to prove `(sec A - tan A)/(sec A + tan A) = (cos^2 A)/(1 + sin A)^2`
Here, we will first solve the LHS.
Now using `sec theta = 1/cos theta` and `tan theta = sin theta/cos theta`, we get
`(sec A - tan A)/(sec A + tan A) = (1/cos A - sin A/cos A)/(1/cos A + sin A/cos A)`
`= ((1 - sin A)/cos A)/((1 + sin A)/cos A)`
`= (1 - sin A)/(1 + sin A)`
Further, multiplying both numerator and denominator by 1 + sin A we get
`(1 - sin A)/(1 + sin A) = ((1 - sin A)/(1 + sin A))((1 + sin A)/(1 = sin A))`
`= ((1 -sin A)(1 + sin A))/(1 + sin A)^2`
`= (1 s sin^2 A)/(1 + sin A)^2`
Now, using the property `cos^2 theta + sin^2 theta = 1`, we get
So,
`(1 - sin^2 A)/(1 + sin A)^2 = cos^2 A/(1 + sin A)^2` = RHS.
Hence proved
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