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Question
`1+((tan^2 theta) cot theta)/(cosec^2 theta) = tan theta`
Solution
LHS = `((1+ tan^2 theta) cot theta)/ (cosec^2 theta) `
=` (sec^2 theta cot theta)/(cosec^2 theta )`
=`(1/cos^2thetaxxcos theta/sin theta)/(1/sin^2 theta)`
=`1/(cos theta sin theta) xx sin^2 theta`
=`sintheta/costheta`
=` tan theta`
=RHS
Hence, LHS = RHS
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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
