If `asin^(-1)x-bcos^(-1)x=c ,` then `asin^(-1)x+bcos^(-1)x` is equal to (a)`0 `(b) `(pia b+c(b-a))/(a+b)` (c)`pi/2` (d) `(pia b+c(a-b))/(a+b)`

Explanation

`asin^(-1)x-bcos^(-1)x=c-(1)`

`b(cos^(-1)x+sin^(-1)x)=pi/2*b-(2)`

Adding equation 1 and 2

`sin^(-1)x(a+b)=(bpi)/2+c`

`sin^(-1)x=((bpi)/2+c)/(a+b)-(3)`

`cosx=pi/2-((bx+2c)/(2(a+b)))`

`=(pia-2c)/(2(a+b))`

`asin^(-1)x+bcos^(-1)x=(xab+c(a-b))/(a+b)`

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