InterviewSolution
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InterviewSolution
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`int tan(x-alpha).tan(x+alpha).tan 2x dx` is equal toA. `ln|(sqrt(sec 2x).sec (x-alpha))/(sec(x-alpha))|+C`B. `ln|(sqrt(sec2x))/(sec(x-alpha).sec(x+alpha))|+C`C. `ln|(sqrt(sec2x).sec (x-alpha))/(sec(x+alpha))|+C`D. `ln|(sec 2x)/(sec(x-alpha).sec(x+alpha))|+C` |
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Answer» Correct Answer - B `tan 2x=tan[(x=alpha)+(x+alpha)]` `=(tan(x-alpha)+tan(x+alpha))/(1-tan (x-alpha).tan(x+alpha))` `rArr" "tan(x-alpha).tan(x+alpha). tan 2x = tan 2x - tan (x-alpha)-tan(x+alpha)` `rArr tan(x-alpha).tan(x-alpha)-tan(x+alpha)]dx` `=(1)/(2)ln|sec 2x|-ln|sec(x-alpha)|-ln |sec(x+alpha)|+C` `=ln|(sqrt(sec2x))/(sec(x-alpha).sec(x+alpha))|+C` |
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