Find the range of `cot^(-1)(2x-x^2)`

1.	Find the range of `cot^(-1)(2x-x^2)`
Answer» Let `theeta = cot^(-1) (2 x - x^(2))`, where `theta in (0, pi)` `rArr cot theta = 2x - x^(2), " where" theta in (0, pi)` `= 1- (1 - 2x + x^(2)), " where " theta in (0, pi)` `= 1 - (1 - x)^(2), " where " theta in (0, pi)` `rArr cot theta le 1, " where " theta in (0, pi)` `rArr (pi)/(4) le theta le pi` `rArr` Range of `f(x) "is " [(pi)/(4), pi)`

Discussion

`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)`
The value of `sin^(-1)(sin12)+sin^(-1)(cos12)=`
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If `x , y , z in [-1,1]`such that `sin^(-1)x+sin^(-1)y+sin^(-1)z=-(3pi)/2,`find the value of `x^2+y^2+z^2dot`A. 1B. 3C. `(3pi^(23))/(4)`D. `3pi^(2)`
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