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