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Question
`1+((tan^2 theta) cot theta)/(cosec^2 theta) = tan theta`
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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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Activity:
L.H.S = `square`
= `square (1 - (sin^2theta)/(tan^2theta))`
= `tan^2theta (1 - square/((sin^2theta)/(cos^2theta)))`
= `tan^2theta (1 - (sin^2theta)/1 xx (cos^2theta)/square)`
= `tan^2theta (1 - square)`
= `tan^2theta xx square` .....[1 – cos2θ = sin2θ]
= R.H.S
Prove that
sec2A – cosec2A = `(2sin^2"A" - 1)/(sin^2"A"*cos^2"A")`
