Question

Show that$\cos ({\sin }^{-1}\left(v\right)+{\cos }^{-1}(v\left)\right)=0$

Short answer

The equation $\cos ({\sin }^{-1}\left(v\right)+{\cos }^{-1}(v\left)\right)=0$can be obtained as

$$\begin{gathered} \cos ({\sin }^{-1}\left(v\right)+{\cos }^{-1}(v\left)\right)=\cos \left(\frac{\mathrm{\pi }}{2}\right) \\ \cos ({\sin }^{-1}\left(v\right)+{\cos }^{-1}(v\left)\right)=0 \end{gathered}$$

Step-by-step solution

Step 1. Given data 

The given equation for verification is

$\cos ({\sin }^{-1}\left(v\right)+{\cos }^{-1}(v\left)\right)=0$

Step 2. Proof

As known${\sin }^{-1}\left(v\right)+{\cos }^{-1}\left(v\right)=\frac{\mathrm{\pi }}{2}\cdots \left(i\right)$

Use equation i in the given equation

$\cos ({\sin }^{-1}\left(v\right)+{\cos }^{-1}(v\left)\right)=\cos \left(\frac{\mathrm{\pi }}{2}\right)$

we know$\cos \left(\frac{\mathrm{\pi }}{2}\right)=0$

So$$\begin{gathered} \cos ({\sin }^{-1}\left(v\right)+{\cos }^{-1}(v\left)\right)=\cos \left(\frac{\mathrm{\pi }}{2}\right) \\ \cos ({\sin }^{-1}\left(v\right)+{\cos }^{-1}(v\left)\right)=0 \end{gathered}$$