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If `cos^(2)alpha+cos^(2)beta=2`, then the value of `tan^(3)alpha+sin^(5)beta` isA. `-1`B. `0`C. `1`D. `(1)/(sqrt(3))` |
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Answer» Correct Answer - b `cos^(2)alpha+cosbeta=2` Put value of `alpha=beta=0^(@)` `rArrcos^(2)0^(@)+cos^(2)0^(@)=2` `rArr(1)^(2)+(1)^(2)=2` `rArr2=2` [it satisfies the equation] `=tan^(2)alpha+sin^(3)beta` `=tan^(2)alpha+sin^(2)beta` `=tan^(2)0^(@)+sin^(2)0^(@)` `=0+0=0` |
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