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If `tanalpha`is equal to the integral solution of the inequality `4x^2-16 x+15 |
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Answer» Correct Answer - D We have `4x^(2)-16x+15lt0` `rArr (3)/(2)ltxlt(5)/(2)` Therefore, the integral solution of `4x^(2)-16x+15lt0` is `x=2`. Thus, `tan alpha=2`. It is given that `cos beta=tan45^(@)=1`. `therefore sin(alpha+beta)sin(alpha-beta)=sin^(2)alpha-sin^(2)beta`. `(1)/(1+cot^(2)alpha)(1-cos^(2)beta)` `=(1)/(1+(1)/(4))-0=(4)/(5)` |
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