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If `sin^6theta+cos^6theta+kcos^2(2theta)=1`, then `k` is |
Answer» `sin^6theta=cos^6theta=(sin^2theta+cos^2theta)(sin^4theta+costheta-sin^2thetacostheta)=1` `(sin^2theta+cos^2theta)^2-3sin^2thetacos^2theta)=1` `=1-3sin^2thetacos^2theta` `=1-3/4*sin^2(2theta)` `=1/4+3/4cos^2 2theta` `3cos^2 2theta+4kcos^2 2 theta=3` `cos^2 2 theta(3+4k)=3` `k=3/4+tan^2theta` `cos^2 2theta*(3+3tan^2 2theta)=3` option d is correct. |
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