3.1 Property 1

2.1 ${\sin ^{ - 1}}x$:
2.2 ${\cos ^{ - 1}}x$:
2.3 ${\tan ^{ - 1}}x$:
2.4 ${\text{cose}}{{\text{c}}^{ - 1}}x$:
2.5 ${\sec^{ - 1}}x$
2.6 ${\cot^{ - 1}}x$
2.7 Summary

3.1 Property 1
3.2 Property 2
3.3 Property 3
3.4 Property 4
3.5 Property 5
3.6 Property 6
3.7 Property 7
3.8 Property 8
3.9 Property 9
3.10 Property 10
3.11 Property 11
3.12 Property 12
3.13 Property 13

(i) ${\sin ^{ - 1}}( - x) = - {\sin ^{ - 1}}x$ where $x \in [ - 1,1]$

Proof: Let $${\sin ^{ - 1}}( - x) = \theta ...(i)$$

$$\begin{equation} \begin{aligned} - x = \sin \theta \\ x = - \sin \theta \\\end{aligned} \end{equation} $$

We know that $$\sin ( - \theta ) = - \sin \theta $$

$$\begin{equation} \begin{aligned} - \theta = {\sin ^{ - 1}}x \\ \theta = - {\sin ^{ - 1}}x \\\end{aligned} \end{equation} $$

From$(i)$, we can write as

$${\sin ^{ - 1}}( - x) = - {\sin ^{ - 1}}x$$

Note: From the graph of ${\sin ^{ - 1}}x$, since it symmetric about origin, if we replace $x$ by $-x$, then the value of $y$ axis shifted from +ve $y$ axis to -ve $y$ axis i.e it becomes $ - {\sin ^{ - 1}}x$.

(ii) $ {\cos ^{ - 1}}( - x) = \pi - {\cos ^{ - 1}}x$ where $x \in [ - 1,1]$

Proof: Let $$\begin{equation} \begin{aligned} {\cos ^{ - 1}}( - x) = \theta ...(i) \\ - x = \cos \theta \\ x = - \cos \theta \\ x = \cos (\pi - \theta ) \\\end{aligned} \end{equation} $$

We know that $(\pi - \theta )$ lies in the second quadrant and value of $cos$ is negative in that quadrant

$${\cos ^{ - 1}}(x) = \pi - \theta $$

Using $(i)$ and then replacing $x$ by $-x$, we get

$${\cos ^{ - 1}}( - x) = \pi - {\cos ^{ - 1}}x$$

(iii) $ {\tan ^{ - 1}}( - x) = - {\tan ^{ - 1}}x,x \in R$

(iv) $ \cos e{c^{ - 1}}( - x) = - \cos e{c^{ - 1}}x$ where $\left| x \right| \geqslant 1$

(v) $ se{c^{ - 1}}( - x) = \pi - se{c^{ - 1}}x$ where $\left| x \right| \geqslant 1$

(vi) ${\cot ^{ - 1}}( - x) = \pi - {\cot ^{ - 1}}x$ where $x \in R$

Question 1. Find the principal value of

(i) ${\cot ^{ - 1}}\left( { - \frac{1}{{\sqrt 3 }}} \right)$

(ii) $\tan \left[ {{{\cos }^{ - 1}}\left( { - \frac{1}{2}} \right) + {{\tan }^{ - 1}}\left( { - \frac{1}{{\sqrt 3 }}} \right)} \right]$

Solution: (i) Let us assume $$y = {\cot ^{ - 1}}\left( { - \frac{1}{{\sqrt 3 }}} \right)$$ We can write $$\begin{equation} \begin{aligned} \cot y = - \frac{1}{{\sqrt 3 }} = \cot \left( {\frac{\pi }{3}} \right) \\ \cot y = \cot \left( {\pi - \frac{\pi }{3}} \right) = \cot \left( {\frac{{2\pi }}{3}} \right) \\\end{aligned} \end{equation} $$ Since we know that the range of ${\cot ^{ - 1}}$ according to the principal value set is $(0,\pi)$ and $$\cot \left( {\frac{{2\pi }}{3}} \right) = - \frac{1}{{\sqrt 3 }}$$ Hence, the principal value of ${\cot ^{ - 1}}\left( { - \frac{1}{{\sqrt 3 }}} \right)$ is ${\frac{{2\pi }}{3}}$

(ii) To solve this, first we have to solve the inverse functions inside bracket and then find its value as $$\begin{equation} \begin{aligned} \tan \left[ {{{\cos }^{ - 1}}\left( { - \frac{1}{2}} \right) + {{\tan }^{ - 1}}\left( { - \frac{1}{{\sqrt 3 }}} \right)} \right] = \tan \left[ {\pi - {{\cos }^{ - 1}}\left( {\frac{1}{2}} \right) - {{\tan }^{ - 1}}\left( {\frac{1}{{\sqrt 3 }}} \right)} \right] \\ = \tan \left[ {\left( {\pi - \frac{\pi }{3}} \right) - \frac{\pi }{6}} \right] = \tan \left[ {\frac{{2\pi }}{3} - \frac{\pi }{6}} \right] \\ = \tan \left[ {\frac{{4\pi - \pi }}{6}} \right] = \tan \left( {\frac{\pi }{2}} \right) = {\text{not defined}} \\\end{aligned} \end{equation} $$

Question 2. Find the principal value of following:

(i) ${\sin ^{ - 1}}\left( {\frac{{ - 1}}{2}} \right)$

(ii) ${\cos ^{ - 1}}\left( {\frac{{\sqrt 3 }}{2}} \right)$

(iii) ${\text{cose}}{{\text{c}}^{ - 1}}(2)$

(iv) ${\tan ^{ - 1}}\left( { - \sqrt 3 } \right)$

(v) ${\sec ^{ - 1}}\left( {\frac{2}{{\sqrt 3 }}} \right)$

$(i)$ Let $${\sin ^{ - 1}}\left( {\frac{{ - 1}}{2}} \right) = y$$

Then, $$\sin y = \frac{{ - 1}}{2} = - \sin \left( {\frac{\pi }{6}} \right) = \sin \left( {\frac{{ - \pi }}{6}} \right)$$

We know that the principal value branch of ${\sin ^{ - 1}}$ is $\left[ { - \frac{\pi }{2},\frac{\pi }{2}} \right]$ and $\sin \left( {\frac{{ - \pi }}{6}} \right) = \frac{{ - 1}}{2}$ so principal value of ${\sin ^{ - 1}}\left( {\frac{{ - 1}}{2}} \right)$ is ${\frac{{ - \pi }}{6}}$

$(ii)$ Let $${\cos ^{ - 1}}\left( {\frac{{\sqrt 3 }}{2}} \right) = y$$

Then, $$\cos y = \left( {\frac{{\sqrt 3 }}{2}} \right) = \cos \left( {\frac{\pi }{6}} \right)$$

We know that the principal value branch of ${\cos ^{ - 1}}$ is $[0,\pi ]$ and $$\cos \left( {\frac{\pi }{6}} \right) = \frac{{\sqrt 3 }}{2}$$

So principal value of ${\cos ^{ - 1}}\left( {\frac{{\sqrt 3 }}{2}} \right)$ is $\frac{\pi }{6}$.

$(iii)$ Let $$\cos e{c^{ - 1}}(2) = y$$

Then, $$\cos ecy = 2 = \cos ec\left( {\frac{\pi }{6}} \right)$$

We know that the principal value branch of $\cos e{c^{ - 1}}$ is $$\left[ { - \frac{\pi }{2},\frac{\pi }{2}} \right] - \{ 0\} $$

So principal value of $\cos e{c^{ - 1}}(2)$ is ${\frac{\pi }{6}}$

$(iv)$ Let $${\tan ^{ - 1}}( - \sqrt 3 ) = y$$

Then, $$\tan y = - \sqrt 3 = - \tan \left( {\frac{\pi }{3}} \right) = \tan \left( { - \frac{\pi }{3}} \right)$$

We know that the principal value branch of ${\tan ^{ - 1}}$ is ${\left( { - \frac{\pi }{2},\frac{\pi }{2}} \right)}$ and $\tan \left( { - \frac{\pi }{3}} \right) = - \sqrt 3 $. So principal value of ${\tan ^{ - 1}}( - \sqrt 3 )$ is ${ - \frac{\pi }{3}}$

$(v)$ Let ${\sec ^{ - 1}}\left( {\frac{2}{{\sqrt 3 }}} \right) = y$

Then, $$\sec y = \frac{2}{{\sqrt 3 }} = \sec \left( {\frac{\pi }{6}} \right)$$

We know that the principal value branch of ${\sec ^{ - 1}}$ is $\left( {0,\pi } \right) - \{ 0\} $ and $\sec \left( {\frac{\pi }{6}} \right) = \frac{2}{{\sqrt 3 }}$

So principal value of ${\sec ^{ - 1}}\left( {\frac{2}{{\sqrt 3 }}} \right)$ is ${\frac{\pi }{6}}$

Question 3. Find the value of $${\tan ^{ - 1}}(1) + {\cos ^{ - 1}}\left( {\frac{{ - 1}}{2}} \right) + {\sin ^{ - 1}}\left( {\frac{{ - 1}}{2}} \right)$$

Solution: Let $${\tan ^{ - 1}}(1) = y$$

Then, $$\begin{equation} \begin{aligned} \tan y = 1 = \tan \left( {\frac{\pi }{4}} \right) \\ {\tan ^{ - 1}}(1) = \frac{\pi }{4} \\\end{aligned} \end{equation} $$

Let $${\cos ^{ - 1}}\left( {\frac{{ - 1}}{2}} \right) = x$$

Then, $$\begin{equation} \begin{aligned} \cos x = \frac{{ - 1}}{2} = - \cos \left( {\frac{\pi }{3}} \right) = \cos \left( {\pi - \frac{\pi }{3}} \right) = \cos \left( {\frac{{2\pi }}{3}} \right) \\ {\cos ^{ - 1}}\left( {\frac{{ - 1}}{2}} \right) = \frac{{2\pi }}{3} \\\end{aligned} \end{equation} $$

Let $${\sin ^{ - 1}}\left( {\frac{{ - 1}}{2}} \right) = z$$

then, $$\begin{equation} \begin{aligned} \sin z = \frac{{ - 1}}{2} = - \sin \left( {\frac{\pi }{6}} \right) = \sin \left( { - \frac{\pi }{6}} \right) \\ {\sin ^{ - 1}}\left( {\frac{{ - 1}}{2}} \right) = - \frac{\pi }{6} \\ \\\end{aligned} \end{equation} $$

Now $$\begin{equation} \begin{aligned} \frac{\pi }{4} + \frac{{2\pi }}{3} - \frac{\pi }{6} \\ = \frac{{3\pi + 8\pi - 2\pi }}{{12}} \\ = \frac{{9\pi }}{{12}} \\ = \frac{{3\pi }}{4} \\\end{aligned} \end{equation} $$