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प्रश्न
`1/((1+tan^2 theta)) + 1/((1+ tan^2 theta))`
उत्तर
LHS=` 1/((1+ tan^2 theta))+1/((1+ cot^2 theta))`
=`1/sec^2 theta + 1/(cosec^2 theta)`
=` cos^2 theta + sin^2 theta`
=1
=RHS
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