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प्रश्न
`(1+tan^2theta)(1+cot^2 theta)=1/((sin^2 theta- sin^4theta))`
उत्तर
LHS= `(1+tan^2theta)(1+cot^2 theta)`
=`sec^2 theta. cosec^2 theta (∵ sec^2 theta - tan^2 theta=1 and cosec^2 - cot^2 theta =1)`
=`1/(cos^2 theta. sin^theta)`
=` 1/((1-sin^2 theta ) sin^2 theta`
=`1/(sin^2theta-sin^4theta)`
==RHS
Hence, LHS = RHS
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