mathematics-class-12-unit-02-chapter-04-inverse-trigonometric-functions-l-4-5-xse4mwczafk

<h3 id="success-stylefont-size25px-inverse-trigonometric-functions-l-4-success"><Success style="font-size:25px"> Inverse Trigonometric Functions L-4 </Success></h3>
<h3 id="primary-stylefont-size24px-inverse-trigonometric-functions-lecture-4-primary"><primary style="font-size:24px"> Inverse trigonometric functions lecture 4 </primary></h3>
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<footer style = "font-size: 18px"> $\rightarrow$  $\rightarrow$ <span style = "color:blue"> Inverse trigonometric functions lecture 4 </span> $\rightarrow$ Conversion between cosine & tangent inverse function $\rightarrow$ Conversion between cosine & tangent inverse function </footer>

### <Success style="font-size:25px"> Inverse Trigonometric Functions L-4 </Success>
### <primary style="font-size:24px"> Conversion between cosine & tangent inverse function </primary>
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- We are looking for something like,

- $ \cos ^{-1} x=\tan ^{-1}(?), \quad|x| \leqslant 1$    
- Let, $ \cos ^{-1} x=\theta, \quad \theta \in[0, \pi]$

- If $x \geqslant 0, \theta \in [0, \frac{\pi}{2}] \subset [-\frac{\pi}{2}, \frac{\pi}{2}].$    
- $ \tan (\cos ^{-1} x)=\tan (\theta)$     
- $ =\frac{\sin (\theta)}{\cos (\theta)}$     
- $ =\frac{\sqrt{1-\cos ^2 \theta}}{x}$

- $\because x= \cos \theta$ ; $\sin^2\theta + \cos^2\theta=1$

- $\Rightarrow \tan (\theta)=\frac{\sqrt{1-x^2}}{x}$   $ [\because \theta \in$ Range set of  $ \tan ^{-1}]$

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<footer style = "font-size: 18px"> $\rightarrow$ Inverse trigonometric functions lecture 4 $\rightarrow$ <span style = "color:blue"> Conversion between cosine & tangent inverse function </span> $\rightarrow$ Conversion between cosine & tangent inverse function $\rightarrow$ Conversion between cosine & tangent inverse function </footer>

<h3 id="success-stylefont-size25px-inverse-trigonometric-functions-l-4-success-1"><Success style="font-size:25px"> Inverse Trigonometric Functions L-4 </Success></h3>
<h3 id="primary-stylefont-size24px-conversion-between-cosine--tangent-inverse-function-primary"><primary style="font-size:24px"> Conversion between cosine & tangent inverse function </primary></h3>
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<p>$ \theta=\cos ^{-1} x=\tan ^{-1} \frac{\sqrt{1-x^2}}{x}, x \geqslant 0$   <br></p>
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<p>If $-1 \leqslant x<0, \theta=\cos ^{-1} x \in\left(\frac{\pi}{2}, \pi\right]$</p>
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<p>$ \tan (\theta)=\frac{\sin \theta}{\cos \theta}=\frac{\sqrt{1-\cos ^2 \theta}}{\cos \theta}=\frac{\sqrt{1-x^2}}{x}$</p>
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<p>$\theta ∉$ Range set of $\tan ^{-1}$</p>
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<p>$\therefore \theta \neq \tan ^{-1} \frac{\sqrt{1-x^2}}{x}$</p>
</li>
<li>
<p>But, $ \tan (\theta)=\tan (\theta-\pi),$</p>
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<p>$ \theta-\pi \in (-\frac{\pi}{2}, 0]$</p>
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<p>$\therefore \theta-\pi \in$ Range set of $ \tan ^{-1}$.</p>
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<footer style = "font-size: 18px">Inverse trigonometric functions lecture 4 $\rightarrow$ Conversion between cosine & tangent inverse function $\rightarrow$ <span style = "color:blue"> Conversion between cosine & tangent inverse function </span> $\rightarrow$ Conversion between cosine & tangent inverse function $\rightarrow$ Conversion from $\tan ^{-1} \text{to} \cos ^{-1}$ </footer>

<h3 id="success-stylefont-size25px-inverse-trigonometric-functions-l-4-success-2"><Success style="font-size:25px"> Inverse Trigonometric Functions L-4 </Success></h3>
<h3 id="primary-stylefont-size24px-conversion-between-cosine--tangent-inverse-function-primary-1"><primary style="font-size:24px"> Conversion between cosine & tangent inverse function </primary></h3>
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<p>$ \tan (\theta)=\tan (\theta-\pi)=\frac{\sqrt{1-x^2}}{x}$</p>
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<p>$\theta-\pi \in$ Range set of $\tan ^{-1}$</p>
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<p>$ \therefore \quad \theta-\pi =\tan ^{-1} \frac{\sqrt{1-x^2}}{x}$</p>
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<p>$ \theta=\pi+\tan ^{-1} \frac{\sqrt{1-x^2}}{x}$   $\quad x<0$</p>
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<p>So,</p>
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<p>$ \cos^{-1} x = \tan^{-1} (\frac{\sqrt{1 - x^2}}{x}) ,~ ~ x \geq 0$</p>
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<p>$ \cos^{-1} x = \pi + \tan^{-1} (\frac{\sqrt{1 - x^2}}{x}) ,~ ~ x < 0$</p>
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<footer style = "font-size: 18px">Conversion between cosine & tangent inverse function $\rightarrow$ Conversion between cosine & tangent inverse function $\rightarrow$ <span style = "color:blue"> Conversion between cosine & tangent inverse function </span> $\rightarrow$ Conversion from $\tan ^{-1} \text{to} \cos ^{-1}$ $\rightarrow$ Conversion from $\tan ^{-1} \text{to} \cos ^{-1}$ </footer>

<h3 id="success-stylefont-size25px-inverse-trigonometric-functions-l-4-success-3"><Success style="font-size:25px"> Inverse Trigonometric Functions L-4 </Success></h3>
<h3 id="primary-stylefont-size24px-conversion-from-tan--1-textto-cos--1-primary"><primary style="font-size:24px"> Conversion from $\tan ^{-1} \text{to} \cos ^{-1}$ </primary></h3>
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<p>Now, we have to make an equation like</p>
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<p>$ \tan ^{-1} x=\cos ^{-1}(?), x \in \mathbb{R}$</p>
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<p>Let, $ x=\tan \theta \Rightarrow \tan ^{-1} x=\theta, \theta \in\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$</p>
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<p>$ \cos \left(\tan ^{-1} x\right)=\cos \theta=\frac{1}{\sec \theta}=\frac{1}{\sqrt{sec^{2}\theta}}$</p>
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<p>$ =\frac{1}{\sqrt{1+\tan ^2 \theta}}$</p>
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<p>$ \cos \theta=\cos \left(\tan ^{-1} x\right)=\left(\frac{1}{\sqrt{1+x^2}}\right)$</p>
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<p>$ \theta=\cos ^{-1} \frac{1}{\sqrt{1+x^2}},$ true only when $\theta \in[0, \pi]$</p>
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<footer style = "font-size: 18px">Conversion between cosine & tangent inverse function $\rightarrow$ Conversion between cosine & tangent inverse function $\rightarrow$ <span style = "color:blue"> Conversion from $\tan ^{-1} \text{to} \cos ^{-1}$ </span> $\rightarrow$ Conversion from $\tan ^{-1} \text{to} \cos ^{-1}$ $\rightarrow$ Conversion from $\tan ^{-1} \text{to} \cos ^{-1}$ </footer>

### <Success style="font-size:25px"> Inverse Trigonometric Functions L-4 </Success>
### <primary style="font-size:24px"> Conversion from $\tan ^{-1} \text{to} \cos ^{-1}$ </primary>
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- We further divide into cases to get respective formulas,
	
- **case I:**  $ x \geqslant 0 $     
- $ \theta=\tan ^{-1} x, \quad \theta \in [0, \frac{\pi}{2})$      
- $ \[0, \frac{\pi}{2}) \subset [0, \pi]$        
- $\leftarrow$ Range set of $\cos ^{-1}$     
      
- $ \theta \in \text { Range set } {\cos ^{-1}}$      
- $ \cos (\theta)=1 / \sqrt{1+x^2},$      
- $ \theta=\tan ^{-1} x=\cos ^{-1} \frac{1}{\sqrt{1+x^2}}, x \geqslant 0$

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<footer style = "font-size: 18px">Conversion between cosine & tangent inverse function $\rightarrow$ Conversion from $\tan ^{-1} \text{to} \cos ^{-1}$ $\rightarrow$ <span style = "color:blue"> Conversion from $\tan ^{-1} \text{to} \cos ^{-1}$ </span> $\rightarrow$ Conversion from $\tan ^{-1} \text{to} \cos ^{-1}$ $\rightarrow$ Conversion formula between $\cot^{-1} ~ and ~ \tan^{-1}$ </footer>

<h3 id="success-stylefont-size25px-inverse-trigonometric-functions-l-4-success-4"><Success style="font-size:25px"> Inverse Trigonometric Functions L-4 </Success></h3>
<h3 id="primary-stylefont-size24px-conversion-from-tan--1-textto-cos--1-primary-1"><primary style="font-size:24px"> Conversion from $\tan ^{-1} \text{to} \cos ^{-1}$ </primary></h3>
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<p><strong>case II:</strong> $ x<0 $</p>
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<p>$ \theta=\tan ^{-1} x, \theta \in {(\frac{-\pi}{2},{0})}$</p>
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<p>$ \theta \notin[0, \pi], \theta \notin \text { Range set of }\cos^{-1}$</p>
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<p>$ \theta+\pi \in\left(\frac{\pi}{2}, \pi\right) \subset[0, \pi]$</p>
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<p>$ \therefore \quad \theta+\pi \in \text { Range set } of \cos ^{-1}$</p>
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<p>$ \cos (\theta+\pi)=-\cos \theta=\frac{-1}{\sqrt{1+x^2}}$</p>
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<p>$ \theta+\pi=\cos ^{-1} \frac{-1}{\sqrt{1+x^2}}$</p>
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<p>$ \theta =-\pi+\cos ^{-1}\frac{-1}{\sqrt{1+x^2}}$</p>
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<footer style = "font-size: 18px">Conversion between cosine & tangent inverse function $\rightarrow$ Conversion between cosine & tangent inverse function $\rightarrow$ <span style = "color:blue"> Conversion from $\tan ^{-1} \text{to} \cos ^{-1}$ </span> $\rightarrow$ Conversion formula between $\cot^{-1} ~ and ~ \tan^{-1}$ $\rightarrow$ Problem </footer>

### <Success style="font-size:25px"> Inverse Trigonometric Functions L-4 </Success>
### <primary style="font-size:24px"> Conversion formula between $\cot^{-1} ~ and ~ \tan^{-1}$ </primary>
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- $\rightarrow \cot ^{-1} x= \tan ^{-1}(1 / x),  x> 0 $

- $\rightarrow  \cot ^{-1} x=\pi+\tan ^{-1}\left(\frac{1}{x}\right),  x<0$

- $\rightarrow  \tan ^{-1} x= \cot ^{-1}\left(\frac{1}{x}\right),  x > 0 $

- $ \rightarrow  \tan ^{-1} x= \cot ^{-1}\left(\frac{1}{x}\right)-\pi,  x<0 $

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<footer style = "font-size: 18px">Conversion between cosine & tangent inverse function $\rightarrow$ Conversion from $\tan ^{-1} \text{to} \cos ^{-1}$ $\rightarrow$ <span style = "color:blue"> Conversion formula between $\cot^{-1} ~ and ~ \tan^{-1}$ </span> $\rightarrow$ Problem $\rightarrow$ Solution </footer>

### <Success style="font-size:25px"> Inverse Trigonometric Functions L-4 </Success>
### <primary style="font-size:24px"> Problem </primary>
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- **<ins>Problem</ins>:**

- Find

- $ \cot \\{\sum_{n=1}^{23} \cot ^{-1} (1+\sum_{k=1}^n 2 k)\\}$

- **<ins>Solution:</ins>**

- $ \cot ^{-1} \left(1+\sum_{k=1}^n 2 k\right)=\cot ^{-1}(1+2 \times(1+2+.......+n))$     
- $ =\cot ^{-1}\left(1+2 \times \frac{n(n+1)}{2}\right)$     
- $ =\cot ^{-1}(1+n(n+1))$

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<footer style = "font-size: 18px">Conversion from $\tan ^{-1} \text{to} \cos ^{-1}$ $\rightarrow$ Conversion formula between $\cot^{-1} ~ and ~ \tan^{-1}$ $\rightarrow$ <span style = "color:blue"> Problem </span> $\rightarrow$ Solution $\rightarrow$ Solution </footer>

### <Success style="font-size:25px"> Inverse Trigonometric Functions L-4 </Success>
### <primary style="font-size:24px"> Solution </primary>
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- = $\tan^{-1}(n+1)-\tan^{-1}(n)$
	
- Let, $ \cot ^{-1}(1+n(n+1))=\theta \in(0, \pi)$ 
	
- $ \cot \theta=1+n(n+1)$    
- $ \tan \theta=\frac{1}{1+n(n+1)}=\frac{(n+1)-n}{1+(\operatorname{n+1}) n}$     
	
- $\because \tan (x-y)=\frac{\tan x-\tan y}{1+\tan x \tan y} $       
- $\Rightarrow \tan \theta =\frac{\tan (\tan ^{-1} (n+1))-\tan (\tan ^{-1} n)}{1+\tan (\tan ^{-1}(n+1)) \tan (\tan ^{-1}n)}$

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<footer style = "font-size: 18px">Conversion formula between $\cot^{-1} ~ and ~ \tan^{-1}$ $\rightarrow$ Problem $\rightarrow$ <span style = "color:blue"> Solution </span> $\rightarrow$ Solution $\rightarrow$ Solution </footer>

<h3 id="success-stylefont-size25px-inverse-trigonometric-functions-l-4-success-5"><Success style="font-size:25px"> Inverse Trigonometric Functions L-4 </Success></h3>
<h3 id="primary-stylefont-size24px-solution-primary"><primary style="font-size:24px"> Solution </primary></h3>
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<ul>
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<p>$ \Rightarrow \tan \theta =\tan (\tan ^{-1}(n+1)-\tan ^{-1} n)$</p>
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<p>($\because n \in N ; n, n+1 > 0 $)</p>
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<p>$ \tan ^{-1}(n+1) \in(0, \pi / 2)$</p>
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<p>$ \tan ^{-1}(n) \in(0, \pi / 2)$</p>
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<p>$ \tan ^{-1}(n+1)-\tan ^{-1} n \in(-\pi / 2, \pi / 2)$  $\leftarrow$ Range set of $\tan^{-1}$</p>
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<p>$ \theta \in(0, \pi / 2)$</p>
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<p>$[\because \theta \in (0,\pi)~$ & $  ~ (\frac{-\pi}{2} , \frac{\pi}{2}) ]$</p>
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<p>$\tan ^{-1}(n+1)-\tan ^{-1} n \in (0, \pi / 2)$</p>
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<footer style = "font-size: 18px">Problem $\rightarrow$ Solution $\rightarrow$ <span style = "color:blue"> Solution </span> $\rightarrow$ Solution $\rightarrow$ Solution </footer>

<h3 id="success-stylefont-size25px-inverse-trigonometric-functions-l-4-success-6"><Success style="font-size:25px"> Inverse Trigonometric Functions L-4 </Success></h3>
<h3 id="primary-stylefont-size24px-solution-primary-1"><primary style="font-size:24px"> Solution </primary></h3>
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<ul>
<li>
<p>$[\because \tan^{-1}$ is increasing function in $(0,\frac{\pi}{2})]$</p>
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<p>$\Rightarrow  \theta=\tan ^{-1}(n+1)-\tan ^{-1}(n)$</p>
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<p>$[n+1>n \Rightarrow \tan^{-1}(n+1)> \tan^{-1}n]$</p>
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<p>$\Rightarrow \cot ^{-1} \left(1+\sum_ {k=1}^n 2 k\right)= \sum_{n=1}^{23}\left(\tan ^{-1} (n+1)-\tan ^{-1} n\right)$</p>
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<p>$ = (\tan ^{-1}  2-\tan ^{-1} 1)+(\tan ^{-1}3 -\tan ^{-1} 2) \cdots +\left(\tan ^{-1}23-\tan ^{-1}22) +\left(\tan ^{-1} 24-\operatorname{tan}^{-1}23\right)\right.$</p>
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<p>$ = \tan ^{-1} 24-\tan ^{-1} 1$</p>
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<footer style = "font-size: 18px">Solution $\rightarrow$ Solution $\rightarrow$ <span style = "color:blue"> Solution </span> $\rightarrow$ Solution $\rightarrow$ Recap </footer>

### <Success style="font-size:25px"> Inverse Trigonometric Functions L-4 </Success>
### <primary style="font-size:24px"> Solution </primary>
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- $ \tan ^{-1} 24-\tan ^{-1} 1$        
- $ =\tan ^{-1} \underbrace{24}_x+\tan ^{-1} \underbrace{(-1)}_y\$        
- $ \tan ^{-1} \frac{24+(-1)}{1-(24)(-1)}=\tan ^{-1} \frac{23}{25}$

- Now, according to the question, 
	
- $\cot \left( \sum_{n=1}^{23} \cot ^{-1}\left(1+\sum_{k=1}^{2 n} 2 k\right)\right)=\cot \left( \tan^{-1}\frac{23}{25} \right) $

- ($\because$ $\tan ^{-1} \frac{23}{25}=\theta$  ,$\Rightarrow$ $\frac{23}{25}=\tan \theta$  , $\Rightarrow$ $\cot \theta=25 / 23$)     
- $=\cot (\theta)$  
- $=25 / 23$

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<footer style = "font-size: 18px">Solution $\rightarrow$ Solution $\rightarrow$ <span style = "color:blue"> Solution </span> $\rightarrow$ Recap $\rightarrow$ Problem </footer>

### <Success style="font-size:25px"> Inverse Trigonometric Functions L-4 </Success>
### <primary style="font-size:24px"> Solution </primary>
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- **<ins>Solution:</ins>**
	
- Observe,

- $\ x-\frac{x^2}{2}+\frac{x^3}{4}-\frac{x^4}{8}+\cdot \cdot \cdot$   
- $=x\left(1-\frac{x}{2}+\frac{x^2}{4}-\frac{x^3}{8}+\cdots\right)$      
- $=x\left(1+\left(-\frac{x}{2}\right)+\left(\frac{-x}{2}\right)^2+\left(-\frac{x}{2}\right)^3\cdots\right.)$      
- $=\frac{x}{1-\left(\frac{-x}{2}\right)} \quad $     
- $=\frac{x}{1+\frac{x}{2}} \quad$    
	
- $\left[ \because |x| < \sqrt{2} \Rightarrow |x| < 2  \Rightarrow |\frac{x}{2}|< 1 \Rightarrow  |\frac{-x}{2}| < 1 \right]$

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<ul>
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<p>And,</p>
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<p>$ \ x^2-\frac{x^4}{2}+\frac{x^6}{4}-\cdots$</p>
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<p>$ =x^2\left(1-\frac{x^2}{2}+\frac{x^4}{4} \cdots \cdot \cdot \cdot\right)$</p>
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<p>$=x^2\left(1+\left(\frac{-x^2}{2}\right)+\left(\frac{-x^2}{2}\right)^2+\cdots \cdot\right)$       $=\frac{x^2}{\left(1-\left(-\frac{x^2}{2}\right)\right)}$</p>
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<p>$=\frac{x^2}{1+\frac{x^2}{2}} \left[\because |x| < \sqrt{2}, x^2 < 2,  \frac{x^2}{2}<1, \left| -\frac{x^2}{2} \right| <1 \right]$</p>
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<h3 id="primary-stylefont-size24px-solution-primary-4"><primary style="font-size:24px"> Solution </primary></h3>
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<li>$\rightarrow \sin ^{-1}\underbrace{\left(\frac{x}{1+\frac{x}{2}}\right)}_\alpha+\cos^{-1}\underbrace{\left(\frac{x^2}{1+\frac{x^2}{2}}\right)} _\beta=\frac{\pi}{2}$</li>
<li>$ \sin ^{-1} \alpha+\cos ^{-1} \beta=\pi / 2$ $~ ~$  (where $\alpha$ & $\beta$ defined above )</li>
<li>$ \sin ^{-1} \alpha=\left(\frac{\pi}{2}-\cos ^{-1} \beta\right)$</li>
<li>$ \alpha=\sin \left(\frac{\pi}{2}-\cos^{-1} \beta \right)$</li>
<li>$ =\cos \left(\cos ^{-1} \beta\right)$</li>
<li>$ =\beta$</li>
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### <Success style="font-size:25px"> Inverse Trigonometric Functions L-4 </Success>
### <primary style="font-size:24px"> Solution </primary>
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- $\Rightarrow \frac{x}{1+\frac{x}{2}}  =\frac{x^2}{1+\frac{x^2}{2}},$      
   
- $\Rightarrow  x\left(1+\frac{x^2}{2}\right)  =x^2\left(1+\frac{x}{2}\right)$      
- $\Rightarrow  x+\frac{x^2}{2} \ =x^2+\frac{x^2}{2}$       
- $ x  (x-1)=0 \quad	\because 0< |x| <\sqrt{2}$       
- $\Rightarrow x=1$

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