ncert-class12-day12-chap2-inverse-trigonometric-functions-theory-part-2-

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### <Primary style = "float:center; text-align:center;"><B>Properties of Inverse Trigonometric Functions</B></Primary>

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<Success style = "float:center; text-align:center; font-size: 22px;"><B>Inverse Trigonometric Functions</B></Success>

#### <Primary style = "float:center; text-align:center; font-size: 22px;"><B>Outline</B></Primary>

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

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

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

- $\operatorname{sin}^{-1} x+\operatorname{cos}^{-1} x=\pi / 2$

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<Success style = "float:center; text-align:center; font-size: 22px;"><B>Inverse Trigonometric Functions</B></Success>

#### <Primary style = "float:center; text-align:center; font-size: 22px;"><B>$\sin (\sin ^{-1} x)=x$</B></Primary>

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If $y=\sin ^{-1} x$, then $x=\sin y$ and if $x=\sin y$, then $y=\sin ^{-1} x$. This is equivalent to

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

$ \sin ^{-1}(\sin x)=x, x \in[\frac{-\pi}{2}, \frac{\pi}{2}]
$

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यदि y = sin⁻¹ x है, तो x = sin y है और यदि x = sin y है, तो y = sin⁻¹ x है। यह निम्न के समतुल्य है:

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

$ \sin ^{-1}(\sin x)=x, x \in[\frac{-\pi}{2}, \frac{\pi}{2}]
$

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<footer>  $\sin (\sin ^{-1} x)$ $\rightarrow$ $\sin^{-1}(-x)$  $\rightarrow$ $\operatorname{Cos}^{-1}(-x)$ $\rightarrow$ More Properties   $\rightarrow$  $\operatorname{Sin}^{-1} x+\operatorname{Cos}^{-1} x$   </footer>

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<Success style = "float:center; text-align:center; font-size: 22px;"><B>Inverse Trigonometric Functions</B></Success>

#### <Primary style = "float:center; text-align:center; font-size: 22px;"><B>$\sin^{-1}(-x) = -\sin^{-1}(x)$</B></Primary>

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Proof:

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

Let $\sin ^{-1}(-x)=y$, i.e., $-x=\sin y$
$\Rightarrow \mathrm{x}=-\sin \mathrm{y}$

Thus,
$$
x=\sin (-y)
$$

Or,

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**सिद्ध:**

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

मान लें $\sin^{-1}(-x) = y$, अर्थात, $-x = \sin y$
$\Rightarrow x = -\sin y$

अतः,

$$
x = \sin (-y)
$$

या,

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<Success style = "float:center; text-align:center; font-size: 22px;"><B>Inverse Trigonometric Functions</B></Success>

#### <Primary style = "float:center; text-align:center; font-size: 22px;"><B>$\sin^{-1}(-x) = -\sin^{-1}(x)$</B></Primary>

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Proof:

Or,
$$
\sin ^{-1}(\mathrm{x})=-\mathrm{y}=-\sin ^{-1}(-\mathrm{x})
$$

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

Similarly, using the same concept following results can be obtained:
- $\operatorname{cosec}^{-1}(-x)=-\operatorname{cosec}^{-1} x,|x| \geq 1$

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

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**सिद्ध: **

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

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

इसी तरह, समान अवधारणा का उपयोग करके निम्नलिखित परिणाम प्राप्त किए जा सकते हैं:

- $\operatorname{cosec}^{-1}(-x) = -\operatorname{cosec}^{-1} x, |x| \geq 1$ (नोट: x का निरपेक्ष मान 1 से बराबर या अधिक होना चाहिए)

- $\ tan^{-1}(-x) = -\ tan^{-1} x, x \in R$ (नोट: x सभी वास्तविक संख्याओं के समुच्चय R का सदस्य हो सकता है)

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<Success style = "float:center; text-align:center; font-size: 22px;"><B>Inverse Trigonometric Functions</B></Success>

#### <Primary style = "float:center; text-align:center; font-size: 22px;"><B>$\operatorname{Cos}^{-1}(-x)=\pi-\operatorname{Cos}^{-1}(x)$</B></Primary>

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

Proof:

Let $\cos ^{-1}(-x)=y$ i.e., $-x=\cos y$
$$
\Rightarrow \mathrm{x}=-\cos \mathrm{y}=\cos (\pi-\mathrm{y})
$$

Thus,
$$
\cos ^{-1}(\mathrm{x})=\pi-\mathrm{y}
$$

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

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<Success style = "float:center; text-align:center; font-size: 22px;"><B>Inverse Trigonometric Functions</B></Success>

#### <Primary style = "float:center; text-align:center; font-size: 22px;"><B>More Properties</B></Primary>

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Similarly using the same concept following results can be obtained:

- $\sec ^{-1}(-\mathrm{x})=\pi-\sec ^{-1} x_r|x| \geq 1$

- $\cot ^{-1}(-\mathrm{x})=\pi-\cot ^{-1} \mathrm{x}, \mathrm{x} \in \mathrm{R}$

- $\operatorname{Tan}^{-1}(-x)=-\operatorname{Tan}^{-1}(x)$

- $\operatorname{Cosec}^{-1}(-x)=-\operatorname{Cosec}^{-1}(x)$

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

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

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<Success style = "float:center; text-align:center; font-size: 22px;"><B>Inverse Trigonometric Functions</B></Success>

#### <Primary style = "float:center; text-align:center; font-size: 22px;"><B>Properties</B></Primary>

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- $\operatorname{Sin}^{-1}(x)=\operatorname{cosec}^{-1}(1 / x), x \in[-1,1]-\\{0\\}$

- $\cos ^{-1}(x)=\sec ^{-1}(1 / x), x \in[-1,1]-\\{0\\}$

- $\operatorname{Tan}^{-1}(x)=\cot ^{-1}(1 / x)$, if $x>0$ (or) $\cot ^{-1}(1 / x)-\pi$, if $x<0$

- $\operatorname{Cot}^{-1}(x)=\tan ^{-1}(1 / x)$, if $x>0$ (or) $\tan ^{-1}(1 / x)+\pi$, if $x<0$

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<Success style = "float:center; text-align:center; font-size: 22px;"><B>Inverse Trigonometric Functions</B></Success>

#### <Primary style = "float:center; text-align:center; font-size: 22px;"><B>$\operatorname{Sin}^{-1} x+\operatorname{Cos}^{-1} x=\pi / 2$</B></Primary>

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Proof: $\sin ^{-1}(x)+\cos ^{-1}(x)=(\pi / 2), x \in[-1,1]$
Let $\sin ^{-1}(x)=y$, i.e., $x=\sin y=\cos ((\pi / 2)-y)$
$\Rightarrow \cos ^{-1}(x)=(\pi / 2)-y=(\pi / 2)-\sin ^{-1}(x)$

Thus,
$$
\sin ^{-1}(x)+\cos ^{-1}(x)=(\pi / 2)
$$

Similarly using the same concept following results can be obtained:
- $\tan ^{-1}(x)+\cot ^{-1}(x)=(\pi / 2), x \in R$
- $\operatorname{cosec}^{-1}(x)+\sec ^{-1}(x)=(\pi / 2),|x| \geq 1$

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मान लें $\sin^{-1}(x) = y$, अर्थात, $x = \sin y = \cos ((\pi / 2) - y)$ (कोज्या का पूरक कोण तंत्रज्ञान)

$\Rightarrow \cos^{-1}(x) = (\pi / 2) - y = (\pi / 2) - \sin^{-1}(x)$

अतः,

$$
\sin^{-1}(x) + \cos^{-1}(x) = (\pi / 2)
$$

- $\ tan^{-1}(x) + \cot^{-1}(x) = (\pi / 2)$, जहाँ x ∈ R

- $\operatorname{cosec}^{-1}(x) + \sec^{-1}(x) = (\pi / 2)$, जहाँ |x| ≥ 1

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<Success style = "float:center; text-align:center; font-size: 22px;"><B>Inverse Trigonometric Functions</B></Success>

#### <Primary style = "float:center; text-align:center; font-size: 22px;"><B>$\operatorname{Tan}^{-1} x+\operatorname{Tan}^{-1} y$</B></Primary>

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(1) If $x, y>0$
$$
\operatorname{Tan}^{-1} x+\operatorname{Tan}^{-1} y=\left\\{\begin{array}{c}
\tan ^{-1}\left(\frac{x+y}{1-x y}\right) x y<1 \\\\
\pi+\tan ^{-1}\left(\frac{x+y}{1-x y}\right) x y>1
\end{array}\right.
$$

Proof: $\operatorname{Tan}^{-1}(x)+\tan ^{-1}(y)=\tan ^{-1}[(x+y) /(1-x y)], x y<1$

Let $\tan ^{-1}(x)=\alpha$ and $\tan ^{-1}(y)=\beta$, i.e., $x=\tan (\alpha)$ and $y=\tan (\beta)$

$\Rightarrow \tan (\alpha+\beta)=(\tan \alpha+\tan \beta) /(1-\tan \alpha \tan \beta)$
Thus,
$(\alpha)+(\beta)=\tan ^{-1}[(x+y) /(1-x y)]$

Therefore,
$$
\tan ^{-1}(\mathrm{x})+\tan ^{-1}(\mathrm{y})=\tan ^{-1}[(\mathrm{x}+\mathrm{y}) /(1-\mathrm{xy})]
$$

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<Success style = "float:center; text-align:center; font-size: 22px;"><B>Inverse Trigonometric Functions</B></Success>

#### <Primary style = "float:center; text-align:center; font-size: 22px;"><B>$\operatorname{Tan}^{-1} x+\operatorname{Tan}^{-1} y$</B></Primary>

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- If $\mathrm{x}, \mathrm{y}<0$

$$
\operatorname{Tan}^{-1} x+\operatorname{Tan}^{-1} y=\left\\{\begin{array}{c}
\tan ^{-1}\left(\frac{x+y}{1-x y}\right) x y<1 \\\\
-\pi+\tan ^{-1}\left(\frac{x+y}{1-x y}\right) x y<1
\end{array}\right.
$$

- $\tan ^{-1}(x)-\tan ^{-1}(y)=\tan^ {-1}[(x-y) /(1+x y)], x y>-1$

- $2 \tan ^{-1}(x)=\tan ^{-1}\left[(2 x) /\left(1-x^2\right)\right],|x|<1$

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<Success style = "float:center; text-align:center; font-size: 22px;"><B>Inverse Trigonometric Functions</B></Success>

#### <Primary style = "float:center; text-align:center; font-size: 22px;"><B>Example</B></Primary>

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**Example 3** Show that

(i) $\sin ^{-1}(2 x \sqrt{1-x^{2}})=2 \sin ^{-1} x,-\frac{1}{\sqrt{2}} \leq x \leq \frac{1}{\sqrt{2}}$

(ii) $\sin ^{-1}(2 x \sqrt{1-x^{2}})=2 \cos ^{-1} x, \frac{1}{\sqrt{2}} \leq x \leq 1$

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<Success style = "float:center; text-align:center; font-size: 22px;"><B>Inverse Trigonometric Functions</B></Success>

#### <Primary style = "float:center; text-align:center; font-size: 22px;"><B>Solution</B></Primary>

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**Solution**

(i) Let $x=\sin \theta$. Then $\sin ^{-1} x=\theta$. We have

$$
\begin{aligned}
\sin ^{-1}(2 x \sqrt{1-x^{2}}) & =\sin ^{-1}(2 \sin \theta \sqrt{1-\sin ^{2} \theta}) \\\\
& =\sin ^{-1}(2 \sin \theta \cos \theta)=\sin ^{-1}(\sin 2 \theta)=2 \theta \\\\
& =2 \sin ^{-1} x
\end{aligned}
$$

(ii) Take $x=\cos \theta$, then proceeding as above, we get, $\sin ^{-1}(2 x \sqrt{1-x^{2}})=2 \cos ^{-1} x$

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<Success style = "float:center; text-align:center; font-size: 22px;"><B>Inverse Trigonometric Functions</B></Success>

#### <Primary style = "float:center; text-align:center; font-size: 22px;"><B>Example</B></Primary>

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**Example 4**

Express $\tan ^{-1} \frac{\cos x}{1-\sin x},-\frac{3 \pi}{2}<x<\frac{\pi}{2}$ in the simplest form.

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<Success style = "float:center; text-align:center; font-size: 22px;"><B>Inverse Trigonometric Functions</B></Success>

#### <Primary style = "float:center; text-align:center; font-size: 22px;"><B>Solution</B></Primary>

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**Solution**

$$
\begin{aligned}
\tan ^{-1}(\frac{\cos x}{1-\sin x}) & =\tan ^{-1}[\frac{\cos ^{2} \frac{x}{2}-\sin ^{2} \frac{x}{2}}{\cos ^{2} \frac{x}{2}+\sin ^{2} \frac{x}{2}-2 \sin \frac{x}{2} \cos \frac{x}{2}}] \\\\
& =\tan ^{-1}[\frac{(\cos \frac{x}{2}+\sin \frac{x}{2})(\cos \frac{x}{2}-\sin \frac{x}{2})}{(\cos \frac{x}{2}-\sin \frac{x}{2})^{2}}] \\\\
& =\tan ^{-1}[\frac{\cos \frac{x}{2}+\sin \frac{x}{2}}{\cos \frac{x}{2}-\sin \frac{x}{2}}]=\tan ^{-1}[\frac{1+\tan \frac{x}{2}}{1-\tan \frac{x}{2}}] \\\\
& =\tan ^{-1}[\tan (\frac{\pi}{4}+\frac{x}{2})]=\frac{\pi}{4}+\frac{x}{2}
\end{aligned}
$$

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<Success style = "float:center; text-align:center; font-size: 22px;"><B>Inverse Trigonometric Functions</B></Success>

#### <Primary style = "float:center; text-align:center; font-size: 22px;"><B>Example</B></Primary>

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**Example 5**

Write $\cot ^{-1}(\frac{1}{\sqrt{x^{2}-1}}), x>1$ in the simplest form.

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<Success style = "float:center; text-align:center; font-size: 22px;"><B>Inverse Trigonometric Functions</B></Success>

#### <Primary style = "float:center; text-align:center; font-size: 22px;"><B>Solution</B></Primary>

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**Solution**

Let $x=\sec \theta$, then $\sqrt{x^{2}-1}=\sqrt{\sec ^{2} \theta-1}=\tan \theta$

Therefore, $\cot ^{-1} \frac{1}{\sqrt{x^{2}-1}}=\cot ^{-1}(\cot \theta)=\theta=\sec ^{-1} x$, which is the simplest form.

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<Success style = "float:center; text-align:center; font-size: 22px;"><B>Inverse Trigonometric Functions</B></Success>

#### <Primary style = "float:center; text-align:center; font-size: 22px;"><B></B></Primary>

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### <Primary>Thank you</Primary>