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प्रश्न
`(1+tan^2theta)(1+cot^2 theta)=1/((sin^2 theta- sin^4theta))`
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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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Solution :
L.H.S. = cotθ + tanθ
= `cosθ/sinθ + sinθ/cosθ`
= `(square + square)/(sinθ xx cosθ)`
= `1/(sinθ xx cosθ)` ............... `square`
= `1/sinθ xx 1/square`
= cosecθ × secθ
L.H.S. = R.H.S
∴ cotθ + tanθ = cosecθ × secθ
