Inverse Sine-Cosine Complementary Relation

Let

$$\sin^{-1}x=A,\qquad \sin^{-1}y=B,\qquad \sin^{-1}z=C$$

Then,

$$A+B+C=\frac{\pi}{2}$$

Hence,

$$C=\frac{\pi}{2}-(A+B)$$

Taking sine on both sides,

$$z=\sin C$$

$$=\sin\left(\frac{\pi}{2}-(A+B)\right)$$

$$=\cos(A+B)$$

Using

$$\cos(A+B)=\cos A\cos B-\sin A\sin B,$$

we get

$$z=\sqrt{1-x^2}\sqrt{1-y^2}-xy$$

Now square both sides:

$$z^2=(1-x^2)(1-y^2)+x^2y^2-2xy\sqrt{(1-x^2)(1-y^2)}$$

Simplifying,

$$z^2=1-x^2-y^2+2x^2y^2-2xy\sqrt{(1-x^2)(1-y^2)}$$

Also,

$$2xyz = 2xy\left(\sqrt{(1-x^2)(1-y^2)}-xy\right)$$

$$= 2xy\sqrt{(1-x^2)(1-y^2)}-2x^2y^2$$

Adding,

$$x^2+y^2+z^2+2xyz$$

$$= x^2+y^2+1-x^2-y^2+2x^2y^2$$

$$\quad-2xy\sqrt{(1-x^2)(1-y^2)}$$

$$\quad+2xy\sqrt{(1-x^2)(1-y^2)}-2x^2y^2$$

$$=1$$

Solution :

Given,

$$\sin^{-1}\left(\frac{3x}{5}\right)+\cos^{-1}\left(\frac{4x}{5}\right)=\frac{\pi}{2}$$

Using identity :

$$\cos^{-1}t=\frac{\pi}{2}-\sin^{-1}t$$

Therefore,

$$\sin^{-1}\left(\frac{3x}{5}\right) + \frac{\pi}{2} - \sin^{-1}\left(\frac{4x}{5}\right) = \frac{\pi}{2}$$

Hence,

$$\sin^{-1}\left(\frac{3x}{5}\right) = \sin^{-1}\left(\frac{4x}{5}\right)$$

Since principal values are equal,

$$\frac{3x}{5}=\frac{4x}{5}$$

$$3x=4x$$

$$x=0$$

Checking :

$$\sin^{-1}(0)+\cos^{-1}(0)$$

$$=0+\frac{\pi}{2}$$

$$=\frac{\pi}{2}$$

Hence verified.

Final Answer :

$$0$$