InterviewSolution
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InterviewSolution
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`tan^(-1)[(cosx)/(1+sinx)]`is equal to`pi/4-x/2,forx in (-pi/2,(3pi)/2)``pi/4-x/2,forx in (-pi/2,pi/2)``pi/4-x/2,forx in (-pi/2,pi/2)``pi/4-x/2,forx in (-(3pi)/2,pi/2)` |
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Answer» `tan^-1((cosx)/(1+sinx))` `=tan^-1((sin(pi/2-x))/(1+cos(pi/2-x)))` `=tan^-1((2sin(pi/4-x/2)cos(pi/4-x/2))/(2cos^2(pi/4-x/2)))` `=tan^-1(tan(pi/4-x/2)` `=pi/4-x/2` Now, we know, range of `tan^-1x` is from `-pi/2` to `pi/2`. `:. -pi/2 lt pi/4-x/2 lt pi/2` `=> -(3pi)/4 lt -x/2 lt pi/4` `=> -pi/2 lt x lt (3pi)/2` So, option `a` is the correct option. |
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