Chapter 7.5, Problem 90AYU

In Problems 87-92, write each trigonometric expression as an algebraic expression containing $U\text{ and }V$. Give the restrictions required on $U\text{ and }V$.

$\sec \left({\tan }^{-1}u{\text{ + cos}}^{-1}v\right)$

To determine

To find: The trigonometric expression $\tan \left({\sin }^{-1}u-{\cos }^{-1}v\right)$ into algebraic expression containing $u$ and $v$ and to give the restrictions on $u$ and $v$.

Answer to Problem 90AYU

Solution:

The trigonometric expression $\tan \left({\sin }^{-1}u-{\cos }^{-1}v\right)$ into algebraic expression containing $u$ and $v$$=\frac{uv-\sqrt{1-v^2}\cdot \sqrt{1-{(u)}^2}}{\sqrt{1-{(u)}^2}\cdot v-u\cdot \sqrt{1-v^2}}$ and the restrictions on $u$ and $v$$=-1\le u\le 1\text{ and }-1\le v\le 1$.

Explanation of Solution

Given:

The trigonometric expression $\tan \left({\sin }^{-1}u-{\cos }^{-1}v\right)$.

Calculation:

Let $\alpha ={\sin }^{-1}u$, $\text{and} \beta ={\cos }^{-1}v$.

$u=\sin \alpha$, $\frac{-\pi }{2}\le \alpha \le \frac{\pi }{2}$, $-1\le u\le 1 \text{and} v=\cos \beta$, $0\le \beta \le \pi$, $-1\le v\le 1$

By Pythagorean identity.

$\cos \alpha =\sqrt{1-{(u)}^2}$, $\sin \beta =\sqrt{1-v^2}$

$\frac{-\pi }{2}\le \alpha \le \frac{\pi }{2}$, $\cos \alpha \ge 0\text{ and }0\le \beta \le \pi$, $\sin \beta \ge 0$

$\tan \left({\sin }^{-1}u-{\cos }^{-1}v\right)=\tan \left(\alpha -\beta \right)$   -----1

By difference formula of $\text{tan}$.

$\tan \left(\alpha -\beta \right)=\frac{\tan \alpha -\tan \beta }{1-\tan \alpha \tan \beta }$   -----2

By quotient identity.

$\tan \left(\alpha -\beta \right)=\frac{\frac{\sin \alpha }{\cos \alpha }-\frac{\sin \beta }{\cos \beta }}{1-\frac{\sin \alpha \sin \beta }{\cos \alpha \cos \beta } }$

Adding up quotients and cancelling the common factor $\cos \alpha \cos \beta$.

$\tan \left(\alpha -\beta \right)=\frac{\frac{\sin \alpha \cos \beta -\sin \beta \cos \alpha }{\cos \alpha \cos \beta }}{\frac{\cos \alpha \cos \beta -\sin \alpha \sin \beta }{\cos \alpha \cos \beta }}$

$\tan \left(\alpha -\beta \right)=\frac{\sin \alpha \cos \beta -\sin \beta \cos \alpha }{\cos \alpha \cos \beta -\sin \alpha \sin \beta }$

$\tan \left(\alpha -\beta \right)=\frac{uv-\sqrt{1-v^2}\cdot \sqrt{1-{(u)}^2}}{\sqrt{1-{(u)}^2}\cdot v-u\cdot \sqrt{1-v^2}}$