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Question
Prove the following:
`1/(1+sin^2theta) + 1/(1+cos^2theta) + 1/(1+sec^2theta) + 1/(1+cosec^2theta) = 2`
Solution
`1/(1+sin^2theta) + 1/(1+cos^2theta) + 1/(1+sec^2theta) + 1/(1+cosec^2theta)`
` = 1/(1+sin^2theta) + 1/(1+cos^2theta) + 1/(1+1/(cos^2theta)) + 1/(1+1/(sin^2theta)) ..........(∵ costheta = 1/sectheta "and" sintheta = 1/(cosectheta))`
`=1/(1+sin^2theta) + 1/(1+cos^2theta) + cos^2 theta/(1+cos^2theta) + sin^2theta/(1+sin^2theta)`
Taking L. C. M
`= ((1+ cos^2theta) + (1+ sin^2theta) + (1 +sin^2theta)(cos^2theta) + (sin^2theta) (1+ cos^2theta))/((1+sin^2theta) (1+cos^2theta))`
`= (1+ cos^2theta +1 + sin^2theta + cos^2theta + sin^2theta cos^2theta +sin^2theta + sin^2thetacos^2theta)/((1+sin^2theta) (1+cos^2theta)) ..(∵ sin^2theta + cos^2theta = 1)`
`= (4+2sin^2thetacos^2theta)/(1+ sin^2theta + cos^2theta + sin^2thetacos^2theta)`
` = (4+2 sin^2thetacos^2theta)/(2+sin^2thetacos^2theta)`
Taking 2 as common factor
`= (2(2+ sin^2thetacos^2theta))/(2+sin^2thetacos^2theta) = 2.`
R. H. S
Hence, proved
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