Maths > Inverse Trigonometric Function > 3.0 Properties
Inverse Trigonometric Function
1.0 Introduction
2.0 Inverse Trigonometric function
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
3.0 Properties
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
3.2 Property 2
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
3.2 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) ${\sin ^{ - 1}}\left( {\frac{1}{x}} \right) = {\text{cose}}{{\text{c}}^{ - 1}}x$ where $x \geqslant 1,x \leqslant - 1$
Proof: We put $${\text{cose}}{{\text{c}}^{ - 1}}(x) = y$$
where $$y \in \left[ {\frac{{ - \pi }}{2},\frac{\pi }{2}} \right] - \{ 0\} $$
$$\begin{equation} \begin{aligned} x = {\text{cosec}}\;y \\ \frac{1}{x} = \frac{1}{{{\text{cosec}}\;y}} = \sin y \\ \frac{1}{x} = \sin y \\ y = {\sin ^{ - 1}}\left( {\frac{1}{x}} \right)...(ii) \\\end{aligned} \end{equation} $$
From $(i)$ and $(ii)$,
$${\sin ^{ - 1}}\left( {\frac{1}{x}} \right) = {\text{cose}}{{\text{c}}^{ - 1}}x$$
Note : the equation $(2)$ must satisfied the condition on $y$ explained above
$$\begin{equation} \begin{aligned} \frac{{ - \pi }}{2} \leqslant {\sin ^{ - 1}}\left( {\frac{1}{x}} \right) \leqslant \frac{\pi }{2} \\ \sin \left( {\frac{{ - \pi }}{2}} \right) \leqslant \frac{1}{x} \leqslant \sin \left( {\frac{\pi }{2}} \right) \\ - 1 \leqslant \frac{1}{x} \leqslant 1 \\\end{aligned} \end{equation} $$
Now take both the inequalities separately
$$\begin{equation} \begin{aligned} - 1 \leqslant \frac{1}{x} \leqslant 1 \\ - 1 \leqslant \frac{1}{x} \\ 0 \leqslant \frac{1}{x} + 1 \\ 0 \leqslant \frac{{1 + x}}{x} \times \frac{x}{x} \\ 0 \leqslant (1 + x)x \\ x \geqslant 0,x \leqslant - 1...(iii) \\\end{aligned} \end{equation} $$
Taking $$\begin{equation} \begin{aligned} \frac{1}{x} \leqslant 1 \\ \frac{1}{x} - 1 \leqslant 0 \\ \frac{{1 - x}}{x} \times \frac{x}{x} \leqslant 0 \\ (1 - x)x \leqslant 0 \\ (x - 1)x \geqslant 0 \\ x \geqslant 1,x \leqslant 0..(iv) \\ y \in \left[ {\frac{{ - \pi }}{2},\frac{\pi }{2}} \right] - \{ 0\} - \operatorname{c} \\\end{aligned} \end{equation} $$
From $(iii)$ and $(iv)$, common region is $x \geqslant 1$ ,$x \leqslant - 1$
(ii) ${\cos ^{ - 1}}\left( {\frac{1}{x}} \right) = {\sec ^{ - 1}}x$ where $x \geqslant 1,x \leqslant - 1$
Proof: We put $$se{c^{ - 1}}(x) = y$$ where $$y \in [0,\pi ] - \left\{ {\frac{\pi }{2}} \right\}$$
$$\begin{equation} \begin{aligned} x = secy \\ \frac{1}{x} = \frac{1}{{secy}} = \cos y...(ii) \\\end{aligned} \end{equation} $$
Note : the equation $(2)$ must satisfied the condition on $y$ explained above
$$\begin{equation} \begin{aligned} 0 \leqslant {\cos ^{ - 1}}\left( {\frac{1}{x}} \right) \leqslant \pi \\ 1 \geqslant \frac{1}{x} \geqslant - 1 \\\end{aligned} \end{equation} $$
Now take both the inequalities separately
$$\begin{equation} \begin{aligned} 1 \geqslant \frac{1}{x} \\ 1 - \frac{1}{x} \geqslant 0 \\ \frac{{x - 1}}{x} \times \frac{x}{x} \leqslant 0 \\ (x - 1)x \geqslant 0 \\ (x - 1)x \geqslant 0 \\ x \geqslant 1,x \leqslant 0..(iii) \\\end{aligned} \end{equation} $$
Taking $$\begin{equation} \begin{aligned} \frac{1}{x} \geqslant - 1 \\ \frac{1}{x} + 1 \geqslant 0 \\ \frac{{1 + x}}{x} \times \frac{x}{x} \geqslant 0 \\ (1 + x)x \geqslant 0 \\ x \geqslant 0,x \leqslant - 1...(iv) \\\end{aligned} \end{equation} $$
From $(iii)$ and $(iv)$, common region is $x \geqslant 1$ ,$x \leqslant - 1$
(iii) \[{\tan ^{ - 1}}\left( {\frac{1}{x}} \right) = \left\{ \begin{gathered} {\cot ^{ - 1}}x;x > 0 \hspace{1em} \\ - \pi + {\cot ^{ - 1}}x;x < 0 \hspace{1em} \\ \end{gathered} \right\}\]
Proof: Put $${\cot ^{ - 1}}x = \theta ..(i)\ where\ \theta \in (0,\pi )$$
Case $(I)$ When $x>0$ i.e $0 < \theta < \frac{\pi }{2}$
$$\begin{equation} \begin{aligned} x = \cot \theta \\ \frac{1}{x} = \frac{1}{{\cot \theta }} = \tan \theta \\ \tan \theta = \frac{1}{x} \\\end{aligned} \end{equation} $$
taking inverse both sides
$${\tan ^{ - 1}}(\tan \theta ) = {\tan ^{ - 1}}\frac{1}{x}$$
Since,$0 < \theta < \frac{\pi }{2}$ ; $${\tan ^{ - 1}}(\tan \theta ) = \theta $$
therefore, we get $$\theta = {\tan ^{ - 1}}\frac{1}{x}$$
From $(i)$, we can write $${\cot ^{ - 1}}x = {\tan ^{ - 1}}\frac{1}{x}$$
Case $(I)$ When $x<0$ i.e $\frac{\pi }{2} < \theta < \pi $
$$\begin{equation} \begin{aligned} x = \cot \theta \\ \frac{1}{x} = \frac{1}{{\cot \theta }} = \tan \theta \\ \tan \theta = \frac{1}{x} \\\end{aligned} \end{equation} $$
taking inverse both sides
$${\tan ^{ - 1}}(\tan \theta ) = {\tan ^{ - 1}}\frac{1}{x}$$
Since, $\frac{\pi }{2} < \theta < \pi $ so ${\tan ^{ - 1}}(\tan \theta ) = \theta - \pi $
Therefore, $$\begin{equation} \begin{aligned} {\tan ^{ - 1}}\frac{1}{x} = \theta - \pi \\ {\tan ^{ - 1}}\frac{1}{x} = {\cot ^{ - 1}}x - \pi \\\end{aligned} \end{equation} $$
