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    				| 1. | Ifsec(theta-alpha),sec(theta+alpha) are in AP , where cos alphane1, then what is the value of sin^(2)theta+cosalpha? | 
| Answer» 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` | |