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प्रश्न
`cot^2 theta - 1/(sin^2 theta ) = -1`a
उत्तर
LHS = `cot^2 theta - 1/ (sin^2 theta)`
= `(cos^2 theta )/(sin^2 theta) - 1/(sin^2 theta)`
=`(cos^2 theta -1)/(sin^2 theta)`
=` (- sin^2 theta )/(sin ^2 theta)`
= -1
= RHS
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