The value of `tan^(-1)tan(-6)`isA. `2pi-6`B. `2pi+6`C. `6-2pi`D. `3pi-6`

The value of `tan^(-1)tan(-6)`isA. `2pi-6`B. `2pi+6`C. `6-2pi`D. `3pi-

1.	The value of `tan^(-1)tan(-6)`isA. `2pi-6`B. `2pi+6`C. `6-2pi`D. `3pi-6`
Answer» We know that `tan^(-1)(tan theta)=theta if -(pi)/(2)le theta le(pi)/(2)` Here`theta =-6` radians which does not lie between However `2pi-6` lies between `-(pi)/(2)and (pi)/(2)` such that `tan(2pi-6)=-tan 6 = tan(-6)` `therefore tan^(-1){tan^(-6)}=tan^(-1){tan(tan(2pi-6))}=2pi-6`

`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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