Question

Writing trigonometric compositions algebraically: Prove each of the following equalities, which rewrite compositions of trigonometric and inverse trigonometric functions as algebraic functions.

$$\begin{gathered} \cos \left({\sin }^{-1}x\right)=\sqrt{1-x^2} \\ \sin \left({\cos }^{-1}x\right)=\sqrt{1-x^2} \\ {\sec }^2\left({\tan }^{-1}x\right)=1+x^2 \\ \tan \left({\sec }^{-1}\mathrm{x}\right)=\left|x\right|\sqrt{1-\frac{1}{x^{\mathit{2}}}} \end{gathered}$$

Short answer

The four equalities:

$$\begin{gathered} \cos \left({\sin }^{-1}x\right)=\sqrt{1-x^2} \\ \sin \left({\cos }^{-1}x\right)=\sqrt{1-x^2} \\ {\sec }^2\left({\tan }^{-1}x\right)=1+x^2 \\ \tan \left({\sec }^{-1}\mathrm{x}\right)=\left|x\right|\sqrt{1-\frac{1}{x^{\mathit{2}}}} \end{gathered}$$

have been proved.

Step-by-step solution

Step 1. Given Information 

We have been given four equalities and we need to prove them.

$$\begin{gathered} \cos \left({\sin }^{-1}x\right)=\sqrt{1-x^2} \\ \sin \left({\cos }^{-1}x\right)=\sqrt{1-x^2} \\ {\sec }^2\left({\tan }^{-1}x\right)=1+x^2 \\ \tan \left({\sec }^{-1}\mathrm{x}\right)=\left|x\right|\sqrt{1-\frac{1}{x^{\mathit{2}}}} \end{gathered}$$

Step 2. Proving the first equality

It is known that ${\sin }^{2}y+{\cos }^{2}y=1$and it can be written as:

${\cos }^{2}y=1-{\sin }^{2}y$.

Now we substitute ${\sin }^{-1}x$for $y$.

role="math" localid="1654264701384" $\begin{array}{rcl}{\cos }^2\left({\sin }^{-1}x\right)& =& 1-{\sin }^2\left({\sin }^{-1}x\right)\\ {\left[\cos \left({\sin }^{-1}x\right)\right]}^2& =& 1-{\left[\sin \left({\sin }^{-1}x\right)\right]}^2\\ {\left[\cos \left({\sin }^{-1}x\right)\right]}^2& =& 1-x^2\\ \cos \left({\sin }^{-1}x\right)& =& \pm \sqrt{1-x^2}\end{array}$

For $-\frac{\mathrm{\pi }}{2}\le y\le \frac{\mathrm{\pi }}{2}$, $\cos y$is always positive. So $\begin{array}{rcl}\cos \left({\sin }^{-1}x\right)& =& \sqrt{1-x^2}\end{array}$.

Thus the first equality is proved.

Step 3. Proving the second equality

It is known that ${\sin }^{2}y+{\cos }^{2}y=1$and it can be written as ${\sin }^{2}y=1-{\cos }^{2}y$.

Now we substitute ${\cos }^{-1}x$for $y$.

$\begin{array}{rcl}{\sin }^2\left({\cos }^{-1}x\right)& =& 1-{\cos }^2\left({\cos }^{-1}x\right)\\ {\left[\sin \left({\cos }^{-1}x\right)\right]}^2& =& 1-{\left[\cos \left({\cos }^{-1}x\right)\right]}^2\\ {\left[\sin \left({\cos }^{-1}x\right)\right]}^2& =& 1-x^2\\ \sin \left({\cos }^{-1}x\right)& =& \pm \sqrt{1-x^2}\end{array}$

For $0\le y\le \mathrm{\pi }$, $\sin y$is always positive. So $\begin{array}{rcl}\sin \left({\cos }^{-1}x\right)& =& \sqrt{1-x^2}\end{array}$.

Thus the second equality is proved.

Step 4. Proving the third equality

It is known that ${\sec }^{2}y-{\tan }^{2}y=1$and it can be written as ${\sec }^{2}y=1+{\tan }^{2}y$.

Now substitute ${\tan }^{-1}x$for $y$

$\begin{array}{rcl}{\sec }^2\left({\tan }^{-1}x\right)& =& 1+{\tan }^2\left({\tan }^{-1}x\right)\\ {\sec }^2\left({\tan }^{-1}x\right)& =& 1+{\left(\tan \left({\tan }^{-1}x\right)\right)}^2\\ {\sec }^2\left({\tan }^{-1}x\right)& =& 1+x^2\end{array}$

Thus the third equality is proved.

Step 5. Proving the fourth equality

It is known that ${\sec }^{2}y-{\tan }^{2}y=1$and it can be written as role="math" localid="1654265261202" ${\tan }^{2}y={\sec }^{2}y-1$.

Now substitute role="math" localid="1654265269848" ${\sec }^{-1}x$for $y$

role="math" localid="1654265385702" $\begin{array}{rcl}{\tan }^2\left({\sec }^{-1}x\right)& =& {\sec }^2\left({\sec }^{-1}x\right)-1\\ {\left(\tan \left({\sec }^{-1}x\right)\right)}^2& =& {\left(\sec \left({\sec }^{-1}x\right)\right)}^2-1\\ {\left(\tan \left({\sec }^{-1}x\right)\right)}^2& =& x^2-1\\ \tan \left({\sec }^{-1}x\right)& =& \sqrt{x^2-1}\\ \tan \left({\sec }^{-1}x\right)& =& \sqrt{x^2\left(1-\frac{1}{x^2}\right)}\\ \tan \left({\sec }^{-1}x\right)& =& \left|x\right|\sqrt{1-\frac{1}{x^2}}\end{array}$

And so the fourth equality is proved.