1.

Ifsec(theta-alpha),sec(theta+alpha) are in AP , where cos alphane1, then what is the value of sin^(2)theta+cosalpha?

Answer»

0
1
`-1`
`1//2`

Solution :`(2)/(costheta)=(cos(theta+ALPHA)+cos(theta-alpha))/(cos(theta+alpha)cos(theta-alpha))=(2costheta.cosalpha)/(cos^(2)theta-sin^(2)alpha)`
`rArrcos^(2)thetacosalpha=cos^(2)theta-sin^(2)alpha`
`rArrsin^(2)alpha=cos^(2)theta(1-cosalpha)`
`rArrcos^(2)theta=(sin^(2)alpha)/(1-cosalpha)=1+cosalpha`
`rArr1-sin^(2)theta=1+cosalpha`
`rArrsin^(2)theta+cosalpha=0`


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