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प्रश्न
`1+((tan^2 theta) cot theta)/(cosec^2 theta) = tan theta`
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उत्तर
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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