Ifasin−1x−b cos−1x=c then the value of asin−1x+bcos−1x=
0
π2
π ab+cb−aa+b
π ab+ca−ba+b
Given asin−1x−bcos−1x=c →1
We know that asin−1x+acos−1x=aπ2 →2
Also bsin-1x+bcos-1x=bπ2 →3
Now equatioins 2-1⇒ Cos−1x=aπ2−ca+b
and equations 3+1⇒sin-1x=bπ2+ca+b
∴ asin−1x+b cos−1x=πab+ca−ba+b