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प्रश्न
`(1 + cot^2 theta ) sin^2 theta =1`
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उत्तर
LHS= `(1+cot^2 theta)sin^2 theta`
=`cosec^2 theta sin^2 theta (∵ cosec^2 theta - cot^2 theta =1)`
=`1/(sin ^2theta)xxsin^2 theta`
=1
Hence, LHS = RHS
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