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Question
Prove the following trigonometric identities
`(1 + tan^2 theta)/(1 + cot^2 theta) = ((1 - tan theta)/(1 - cot theta))^2 = tan^2 theta`
Solution
We have to prove `(1 + tan^2 theta)/(1 + cot^2 theta) = ((1 - tan theta)/(1 - cot theta))^2 = tan^2 theta`
Consider the expression
L.H.S
`(1 + tan^2 theta)/(1 + cot^2 theta) = (1 + tan^2 theta)/(1 + 1/(tan^2 theta))`
= `(1 +tan^2 theta)/((tan^2 theta + 1)/tan^2 theta)`
`= tan^2 theta (1 + tan^2 theta)/(1 + tan^2 theta)`
`= tan^2 theta`
= R.H.S
Again, we have
L.H.S
`((1 - tan theta)/(1 - cot theta))^2 = ((1 - tan theta)/(1 - 1/(tan theta)))^2`
`= ((1 - tan theta)/((tan theta - 1)/tan theta))^2`
`=[(tantheta(1-tantheta))/-(1-tantheta)]^2`
`=(-tantheta)^2=tan^2theta`
= R.H.S
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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
