Question

Prove the following:

`1/(1+sin^2theta) + 1/(1+cos^2theta) + 1/(1+sec^2theta) + 1/(1+cosec^2theta) = 2`

Answer 1

`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