Inverse Trigonometric Functions

Prove that: $\;$ $\tan^{-1} \left(\dfrac{1}{7}\right) + \tan^{-1} \left(\dfrac{1}{13}\right) = \tan^{-1} \left(\dfrac{2}{9}\right)$

$\tan^{-1} \left(x\right) + \tan^{-1} \left(y\right) = \tan^{-1} \left(\dfrac{x + y}{1 - x \cdot y}\right), \;\; x \cdot y < 1$

$\begin{aligned} \therefore \; \tan^{-1} \left(\dfrac{1}{7}\right) + \tan^{-1} \left(\dfrac{1}{13}\right) & = \tan^{-1} \left(\dfrac{\dfrac{1}{7} + \dfrac{1}{13}}{1 - \dfrac{1}{7} \times \dfrac{1}{13}}\right) \\\\ & = \tan^{-1} \left(\dfrac{20}{90}\right) \\\\ & = \tan^{-1} \left(\dfrac{2}{9}\right) \end{aligned}$

Hence proved.

Inverse Trigonometric Functions

Find the value of: $\;$ $\sin^{-1} \left[\sin \left(\dfrac{5 \pi}{6}\right)\right] + \cos^{-1} \left[\cos \left(\dfrac{5 \pi}{3}\right) \right] + \tan^{-1} \left[\tan \left(\dfrac{7 \pi}{3}\right)\right]$

$\begin{aligned} \sin^{-1} \left[\sin \left(\dfrac{5 \pi}{6}\right)\right] & = \sin^{-1} \left[\sin \left(\pi - \dfrac{\pi}{6}\right)\right] & \hspace{1cm} \left\{\dfrac{5 \pi}{6} \notin \left[- \dfrac{\pi}{2}, \dfrac{\pi}{2}\right] \right\} \\\\ & = \sin^{-1} \left[\sin \left(\dfrac{\pi}{6}\right)\right] & \\\\ & = \dfrac{\pi}{6} \hspace{1cm} & \hspace{1cm} \left\{\dfrac{\pi}{6} \in \left[- \dfrac{\pi}{2}, \dfrac{\pi}{2}\right] \right\} \end{aligned}$

$\begin{aligned} \cos^{-1} \left[\cos \left(\dfrac{5 \pi}{3}\right)\right] & = \cos^{-1} \left[\cos \left(2 \pi - \dfrac{\pi}{3}\right)\right] & \hspace{1cm} \left\{\dfrac{5 \pi}{3} \notin \left[0, \pi\right] \right\} \\\\ & = \cos^{-1} \left[\cos \left(\dfrac{\pi}{3}\right)\right] & \\\\ & = \dfrac{\pi}{3} \hspace{1cm} & \hspace{1cm} \left\{\dfrac{\pi}{3} \in \left[0, \pi\right] \right\} \end{aligned}$

$\begin{aligned} \tan^{-1} \left[\tan \left(\dfrac{7 \pi}{3}\right)\right] & = \tan^{-1} \left[\tan \left(2 \pi + \dfrac{\pi}{3}\right)\right] & \hspace{1cm} \left\{\dfrac{7 \pi}{3} \notin \left(- \dfrac{\pi}{2}, \dfrac{\pi}{2}\right) \right\} \\\\ & = \tan^{-1} \left[\tan \left(\dfrac{\pi}{3}\right)\right] & \\\\ & = \dfrac{\pi}{3} \hspace{1cm} & \hspace{1cm} \left\{\dfrac{\pi}{3} \in \left(- \dfrac{\pi}{2}, \dfrac{\pi}{2}\right) \right\} \end{aligned}$

$\therefore \; \sin^{-1} \left[\sin \left(\dfrac{5 \pi}{6}\right)\right] + \cos^{-1} \left[\cos \left(\dfrac{5 \pi}{3}\right) \right] + \tan^{-1} \left[\tan \left(\dfrac{7 \pi}{3}\right)\right]$

$= \dfrac{\pi}{6} + \dfrac{\pi}{3} + \dfrac{\pi}{3}$

$= \dfrac{5 \pi}{6}$

Inverse Trigonometric Functions

Find the value of: $\;$ $\cos \left[\sin^{-1} \left(- \dfrac{4}{5}\right)\right] + \sin \left[\tan^{-1} \left(\dfrac{3}{4}\right)\right] + \cos \left[\csc^{-1} \left(\dfrac{5}{3}\right)\right]$

Let $\;$ $\sin^{-1} \left(- \dfrac{4}{5}\right) = - \sin^{-1} \left(\dfrac{4}{5}\right) = -\alpha, \hspace{1cm} -\alpha \in \left[- \dfrac{\pi}{2}, \dfrac{\pi}{2}\right]$

Then, $\;$ $\sin \left(\alpha\right) = \dfrac{4}{5}$ $\;\;\; \cdots \; (1)$

Let $\;$ $\tan^{-1} \left(\dfrac{3}{4}\right) = \beta, \hspace{1cm} \beta \in \left(- \dfrac{\pi}{2}, \dfrac{\pi}{2}\right)$

Then, $\;$ $\tan \left(\beta\right) = \dfrac{3}{4}$ $\;\;\; \cdots \; (2)$

Let $\;$ $\csc^{-1} \left(\dfrac{5}{3}\right) = \gamma, \hspace{1cm} \gamma \in \left[- \dfrac{\pi}{2}, \dfrac{\pi}{2}\right] - \left\{0\right\}$

Then, $\;$ $\csc \left(\gamma\right) = \dfrac{5}{3}$ $\implies$ $\sin \left(\gamma\right) = \dfrac{3}{5}$ $\;\;\; \cdots \; (3)$

From equation $(1)$, $\;$ $\cos \left(\alpha\right) = \sqrt{1 - \sin^2 \left(\alpha\right)} = \sqrt{1 - \dfrac{16}{25}} = \sqrt{\dfrac{9}{25}} = \dfrac{3}{5}$ $\;\; \cdots \; (4a)$

From equation $(2)$, $\;$ $\sin \left(\beta\right) = \dfrac{3}{\sqrt{3^2 + 4^2}} = \dfrac{3}{\sqrt{25}} = \dfrac{3}{5}$ $\;\;\; \cdots \; (4b)$

From equation $(3)$, $\;$ $\cos \left(\gamma\right) = \sqrt{1 - \sin^2\left(\gamma\right)} = \sqrt{1 - \dfrac{9}{25}} = \sqrt{\dfrac{16}{25}} = \dfrac{4}{5}$ $\;\;\; \cdots \; (4c)$

$\therefore \;$ $\cos \left[\sin^{-1} \left(- \dfrac{4}{5}\right)\right] + \sin \left[\tan^{-1} \left(\dfrac{3}{4}\right)\right] + \cos \left[\csc^{-1} \left(\dfrac{5}{3}\right)\right]$

$= \cos \left(- \alpha\right) + \sin \left(\beta\right) + \cos \left(\gamma\right)$

$= \cos \left(\alpha\right) + \sin \left(\beta\right) + \cos \left(\gamma\right)$ $\hspace{1cm} \left[\because \; \cos \left(- \theta\right) = \cos \left(\theta\right)\right]$

$= \dfrac{3}{5} + \dfrac{3}{5} + \dfrac{4}{5}$ $\hspace{1cm}$ [from equations $(4a), \; (4b), \; (4c)$]

$= \dfrac{10}{5} = 2$

Inverse Trigonometric Functions

Find the value of: $\;$ $\cot^{-1} \left(1\right) + 3 \sin^{-1} \left(\dfrac{1}{2}\right) - \csc^{-1} \left(- 2\right) - 3 \tan^{-1} \left(\dfrac{1}{\sqrt{3}}\right)$

$\cot^{-1} \left(1\right) = \cot^{-1} \left[\cot \left(\dfrac{\pi}{4}\right)\right] = \dfrac{\pi}{4}, \hspace{1cm} \dfrac{\pi}{4} \in \left(0, \pi\right)$

$\sin^{-1} \left(\dfrac{1}{2}\right) = \sin^{-1} \left[\sin \left(\dfrac{\pi}{6}\right)\right] = \dfrac{\pi}{6}, \hspace{1cm} \dfrac{\pi}{6} \in \left[- \dfrac{\pi}{2}, \dfrac{\pi}{2}\right]$

$\csc^{-1} \left(- 2\right) = \csc^{-1} \left[- \csc \left(\dfrac{\pi}{6}\right)\right] = \csc^{-1} \left[\csc \left(- \dfrac{\pi}{6}\right)\right] = - \dfrac{\pi}{6}$,

where $\;$ $- \dfrac{\pi}{6} \in \left[- \dfrac{\pi}{2}, \dfrac{\pi}{2}\right] - \left\{0\right\}$

$\tan^{-1} \left(\dfrac{1}{\sqrt{3}}\right) = \tan^{-1} \left[\tan \left(\dfrac{\pi}{6}\right)\right] = \dfrac{\pi}{6}, \hspace{1cm} \dfrac{\pi}{6} \in \left(- \dfrac{\pi}{2}, \dfrac{\pi}{2}\right)$

$\therefore \;$ $\cot^{-1} \left(1\right) + 3 \sin^{-1} \left(\dfrac{1}{2}\right) - \csc^{-1} \left(- 2\right) - 3 \tan^{-1} \left(\dfrac{1}{\sqrt{3}}\right)$

$= \dfrac{\pi}{4} + 3 \times \dfrac{\pi}{6} + \dfrac{\pi}{6} - 3 \times \dfrac{\pi}{6}$

$= \dfrac{\pi}{4} + \dfrac{\pi}{6} = \dfrac{5 \pi}{12}$

Inverse Trigonometric Functions

Find the value of: $\;$ $\sin^{-1} \left(\dfrac{\sqrt{3}}{2}\right) - \cos^{-1} \left(- \dfrac{\sqrt{3}}{2}\right) + 2 \tan^{-1} \left(1\right)$

$\sin^{-1} \left(\dfrac{\sqrt{3}}{2}\right) = \sin^{-1} \left[\sin \left(\dfrac{\pi}{3}\right)\right] = \dfrac{\pi}{3}, \hspace{1cm} \dfrac{\pi}{3} \in \left[- \dfrac{\pi}{2}, \dfrac{\pi}{2}\right]$

$\cos^{-1} \left(- \dfrac{\sqrt{3}}{2}\right) = \pi - \cos^{-1} \left(\dfrac{\sqrt{3}}{2}\right) = \pi - \dfrac{\pi}{6} = \dfrac{5 \pi}{6}, \hspace{1cm} \dfrac{5 \pi}{6} \in \left[0, \pi\right]$

$\tan^{-1} \left(1\right) = \tan^{-1} \left[\tan \left(\dfrac{\pi}{4}\right)\right] = \dfrac{\pi}{4}, \hspace{1cm} \dfrac{\pi}{4} \in \left(- \dfrac{\pi}{2}, \dfrac{\pi}{2}\right)$

$\begin{aligned} \therefore \; \sin^{-1} \left(\dfrac{\sqrt{3}}{2}\right) - \cos^{-1} \left(- \dfrac{\sqrt{3}}{2}\right) + 2 \tan^{-1} \left(1\right) & = \dfrac{\pi}{3} - \dfrac{5 \pi}{6} + 2 \times \dfrac{\pi}{4} \\\\ & = \dfrac{\pi}{3} - \dfrac{5 \pi}{6} + \dfrac{\pi}{2} \\\\ & = \dfrac{2 \pi - 5 \pi + 3 \pi}{6} \\\\ & = 0 \end{aligned}$

Inverse Trigonometric Functions

Evaluate: $\;$ $\sin \left[3 \sin^{-1} \left(\dfrac{1}{3}\right)\right]$

Let $\sin^{-1} \left(\dfrac{1}{3}\right) = \theta \;\;\; \text{ where } \theta \in \left[- \dfrac{\pi}{2}, \dfrac{\pi}{2}\right]$

Then, $\sin \left(\theta\right) = \dfrac{1}{3}$

$\begin{aligned} \therefore \; \sin \left[3 \sin^{-1} \left(\dfrac{1}{3}\right)\right] & = \sin \left(3 \theta\right) \\\\ & = 3 \sin \theta - 4 \sin^3 \theta \\\\ & = 3 \times \dfrac{1}{3} - 4 \times \left(\dfrac{1}{3}\right)^3 \\\\ & = 1 - \dfrac{4}{27} \\\\ & = \dfrac{23}{27} \end{aligned}$

Inverse Trigonometric Functions

Evaluate: $\;$ $\tan^2 \left[\dfrac{1}{2} \sin^{-1} \left(\dfrac{2}{3}\right)\right]$

Let $\sin^{-1} \left(\dfrac{2}{3}\right) = \theta \;\;\; \text{ where } \theta \in \left[- \dfrac{\pi}{2}, \dfrac{\pi}{2}\right]$

Then, $\sin \left(\theta\right) = \dfrac{2}{3}$

$ \therefore \; \tan^2 \left[\dfrac{1}{2} \sin^{-1} \left(\dfrac{2}{3}\right)\right] = \tan^2 \left(\dfrac{\theta}{2}\right)$

Now, $\;$ $\sin \left(\theta\right) = \dfrac{2 \tan \left(\dfrac{\theta}{2}\right)}{1 + \tan^2 \left(\dfrac{\theta}{2}\right)}$

i.e. $\;$ $\dfrac{2 \tan \left(\dfrac{\theta}{2}\right)}{1 + \tan^2 \left(\dfrac{\theta}{2}\right)} = \dfrac{2}{3}$

i.e. $\;$ $\tan^2 \left(\dfrac{\theta}{2}\right) - 3 \tan \left(\dfrac{\theta}{2}\right) + 1 = 0$

i.e. $\;$ $\tan \left(\dfrac{\theta}{2}\right) = \dfrac{3 \pm \sqrt{9 - 4}}{2}$

i.e. $\;$ $\tan \left(\dfrac{\theta}{2}\right) = \dfrac{3 + \sqrt{5}}{2}$ $\;$ OR $\;$ $\tan \left(\dfrac{\theta}{2}\right) = \dfrac{3 - \sqrt{5}}{2}$

$\therefore \;$ $\tan^2 \left(\dfrac{\theta}{2}\right) = \left(\dfrac{3 + \sqrt{5}}{2}\right)^2$ $\;$ OR $\;$ $\tan^2 \left(\dfrac{\theta}{2}\right) = \left(\dfrac{3 - \sqrt{5}}{2}\right)^2$

i.e. $\;$ $\tan^2 \left(\dfrac{\theta}{2}\right) = \dfrac{14 + 6 \sqrt{5}}{4}$ $\;$ OR $\;$ $\tan^2 \left(\dfrac{\theta}{2}\right) = \dfrac{14 - 6 \sqrt{5}}{4}$

i.e. $\;$ $\tan^2 \left(\dfrac{\theta}{2}\right) = \dfrac{7 \pm 3 \sqrt{5}}{2}$

$\because \;$ $\dfrac{7 + 3 \sqrt{5}}{2} \; \notin \; \left[-1, 1\right]$ $\;$ and $\;$ $\dfrac{7 - 3 \sqrt{5}}{2} \; \in \; \left[-1, 1\right]$,

$\therefore \;$ $\tan^2 \left(\dfrac{\theta}{2}\right) = \dfrac{7 - 3 \sqrt{5}}{2}$

i.e. $\;$ $\tan^2 \left[\dfrac{1}{2} \sin^{-1} \left(\dfrac{2}{3}\right)\right] = \dfrac{7 - 3 \sqrt{5}}{2}$