mathematics-class-12-unit-02-chapter-05-inverse-trigonometric-functions-l-5-5-wu-vsicwqwi

### <Success style="font-size:25px"> Inverse Trigonometric Functions L-5 </Success>
### <primary style="font-size:24px"> Problem </primary>
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- **<ins>Problem:1</ins>**

- Prove that,
- $\cos \left[\tan ^{-1}\left[\sin \left(\cot ^{-1} x\right)\right]\right]=\sqrt{\frac{x^2+1}{x^2+2}}$ .

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

- Let, $\cot ^{-1} x=\theta \in(0, \pi) \Rightarrow$ $ x=\cot (\theta)$ , $\tan \theta=\frac{1}{x}  $    
- $\sin \theta =\sqrt{\sin ^2 \theta} \quad \because \theta \in(0,\pi) \Rightarrow \sin \theta > 0$

- $=\sqrt{1-\cos ^2 \theta}  $ 
- $=\sqrt{1-\frac{1}{\sec ^2 \theta}}$   
- $=\sqrt{1-\frac{1}{1+\tan ^2 \theta}}$  
- $=\sqrt{\frac{\tan ^2 \theta}{1+\tan ^2 \theta}}$

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<footer style = "font-size: 18px"> $\rightarrow$ Inverse trigonometric functions problem solving lecture 5 $\rightarrow$ <span style = "color:blue"> Problem </span> $\rightarrow$ Solution $\rightarrow$ Solution </footer>

### <Success style="font-size:25px"> Inverse Trigonometric Functions L-5 </Success>
### <primary style="font-size:24px"> Solution </primary>
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- $\sin \theta=\sqrt{\frac{\frac{1}{x^2}}{1+\frac{1}{x^2}}}=\sqrt{\frac{1}{1+x^2}}$ $\Rightarrow \sin (\cot^{-1})=\sqrt{\frac{1}{1+x^2}}$

- Now, substituting in the expression we get,
- $\cos \left[\left(\tan ^{-1} \frac{1}{\sqrt{1+x^2}}\right)\right]$

- Let, $\tan ^{-1}\left(\frac{1}{\sqrt{1+x^2}}\right)=\phi \in\left(0, \frac{\pi}{2}\right) \because \theta > 0$

- $\tan \phi=\frac{1}{\sqrt{1+x^2}} $ , again substitute the expression of $\tan^{-1}(\frac{1}{1+x^{2}})$ we get,

- $\cos \phi=\sqrt{\cos ^2 \phi}=\frac{1}{\sqrt{\sec ^2 \phi}}=\frac{1}{\sqrt{1+\tan ^2 \phi}} \cdots (a) $

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<footer style = "font-size: 18px">Inverse trigonometric functions problem solving lecture 5 $\rightarrow$ Problem $\rightarrow$ <span style = "color:blue"> Solution </span> $\rightarrow$ Solution $\rightarrow$ Problem </footer>

### <Success style="font-size:25px"> Inverse Trigonometric Functions L-5 </Success>
### <primary style="font-size:24px"> Solution </primary>
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- We know, 
- $\tan \phi=\frac{1}{\sqrt{1+x^2}}$  on substitution in $\cdots$ (a) we get,

- $\cos \phi=\frac{1}{\sqrt{1+\tan ^2 \phi}}$ $=\frac{1}{\sqrt{1+\frac{1}{1+x^2}}}$

- $=\sqrt{\frac{1+x^2}{2+x^2}}$

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

<h3 id="success-stylefont-size25px-inverse-trigonometric-functions-l-5-success"><Success style="font-size:25px"> Inverse Trigonometric Functions L-5 </Success></h3>
<h3 id="primary-stylefont-size24px-solution-primary"><primary style="font-size:24px"> Solution </primary></h3>
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<p><strong><ins>Solution:</ins></strong></p>
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<p>$\theta \in[0,4 \pi]$ such that</p>
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<p>$\cos ^{-1}(\cos \theta)=\frac{10-\theta}{10}$</p>
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<p>${\theta \in[0,4 \pi]}$</p>
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<p>Divide the region $[0,4\pi]$ into 4 regions,</p>
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<img src="https://imagedelivery.net/YfdZ0yYuJi8R0IouXWrMsA/4bb3a89e-27e4-4da2-c6a8-bc255cee2800/public" alt="">

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

<h3 id="success-stylefont-size25px-inverse-trigonometric-functions-l-5-success-2"><Success style="font-size:25px"> Inverse Trigonometric Functions L-5 </Success></h3>
<h3 id="primary-stylefont-size24px-solution-primary-2"><primary style="font-size:24px"> Solution </primary></h3>
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<p>$\quad \theta \in(\pi, 2 \pi] \Leftrightarrow(2\pi-\theta) \in[0, \pi] ⊆ $ Range set of $\cos ^{-1}$</p>
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<p>$\cos ^{-1}(\cos \theta)≠\theta$</p>
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<p>{$\because \cos ^{-1}(\cos \theta) \in [0,\pi]$ and $\theta \in[\pi, 2\pi]$ }</p>
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<p>$\cos ^{-1}(\cos \theta)=x , x\in[0, \pi]$</p>
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<p>$\cos \theta=\cos x$</p>
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<p>$\cos (\underbrace{2 \pi-\theta}_\phi)=(\cos \theta)$</p>
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<footer style = "font-size: 18px">Solution $\rightarrow$ Solution $\rightarrow$ <span style = "color:blue"> Solution </span> $\rightarrow$ Solution $\rightarrow$ Solution </footer>

### <Success style="font-size:25px"> Inverse Trigonometric Functions L-5 </Success>
### <primary style="font-size:24px"> Solution </primary>
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- If $\theta \in(2 \pi, 3 \pi]$

- $\theta-2 \pi \in{(0, \pi]} ⊆ $ Range set of $\cos ^{-1}$   
- $\cos (\theta-2 \pi)=\cos \theta$  
- $\theta-2\pi=\cos ^{-1}(\cos \theta)$

- If $\theta \in(3 \pi, 4 \pi]$  
- $(4 \pi-\theta) \in[0, \pi) ⊆ $ Range set of $\cos ^{-1}$  
- $\cos (4 \pi-\theta)=\cos \theta$   
- $ 4\pi-\theta=\cos ^{-1}(\cos \theta)$

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

<h3 id="success-stylefont-size25px-inverse-trigonometric-functions-l-5-success-4"><Success style="font-size:25px"> Inverse Trigonometric Functions L-5 </Success></h3>
<h3 id="primary-stylefont-size24px-solution-primary-4"><primary style="font-size:24px"> Solution </primary></h3>
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<li>$f(\theta)=\cos ^{-1}(\cos \theta)= \begin{cases}\theta,  \theta \in[0, \pi] \\ 2 \pi-\theta,  \theta \in[\pi, 2 \pi] \\ \theta-2 \pi,  \theta \in[2 \pi, 3 \pi] \\ 4 \pi-\theta,  \theta \in[3 \pi, 4 \pi]\end{cases}$</li>
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<img src="https://imagedelivery.net/YfdZ0yYuJi8R0IouXWrMsA/33bd9754-a8b5-4af9-e4d2-8d2d26083a00/public" alt="">

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

### <Success style="font-size:25px"> Inverse Trigonometric Functions L-5 </Success>
### <primary style="font-size:24px"> Problem </primary>
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- <b><ins>Problem:3</ins></b>  Solve the following equation for $x$  
- $(\tan ^{-1} 2 x)$+ $(\tan ^{-1} 3 x)=\frac{\pi}{4},   x>0 . $

- <b><ins>Solution:</ins></b>

- $\tan ^{-1} 2 x =(\frac{\pi}{4})-\tan ^{-1} 3 x$  
- $=\tan ^{-1} 1-\tan ^{-1} 3 x$  
- $\tan ^{-1} 1+\tan ^{-1}(-3 x)$  
- $\tan ^{-1}(2 x)=\tan ^{-1}(\frac{1-3 x}{1+3 x})$  
- $\tan ^{-1} a =\tan ^{-1} b$  
- $ a = b$

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

<h3 id="success-stylefont-size25px-inverse-trigonometric-functions-l-5-success-6"><Success style="font-size:25px"> Inverse Trigonometric Functions L-5 </Success></h3>
<h3 id="primary-stylefont-size24px-solution-primary-6"><primary style="font-size:24px"> Solution </primary></h3>
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<ul>
<li>$2 x=\frac{1-3 x}{1+3 x}$</li>
<li>$2 x(1+3 x)=1-3 x$</li>
<li>$6 x^2+5 x-1=0 $</li>
<li>$(6 x-1)(x+1)=0 $</li>
<li>$x=\frac{1}{6}, -1$</li>
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<footer style = "font-size: 18px">Problem $\rightarrow$ Solution $\rightarrow$ <span style = "color:blue"> Solution </span> $\rightarrow$ Problem $\rightarrow$ Solution </footer>

<h3 id="success-stylefont-size25px-inverse-trigonometric-functions-l-5-success-8"><Success style="font-size:25px"> Inverse Trigonometric Functions L-5 </Success></h3>
<h3 id="primary-stylefont-size24px-solution-primary-7"><primary style="font-size:24px"> Solution </primary></h3>
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<p><b><ins>Solution:</ins></b></p>
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<p>$y=\frac{x+z}{2}, \quad y-x=z-y $</p>
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<p>$ \underbrace{\tan ^{-1} y-\tan ^{-1} x}_{\theta \geqslant 0}=\tan ^{-1} z-\tan ^{-1} y$</p>
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<p>$2\theta<\pi$</p>
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<p>$0 \leqslant \theta < \frac{\pi}{2}$</p>
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<footer style = "font-size: 18px">Solution $\rightarrow$ Problem $\rightarrow$ <span style = "color:blue"> Solution </span> $\rightarrow$ Solution $\rightarrow$ Solution </footer>

### <Success style="font-size:25px"> Inverse Trigonometric Functions L-5 </Success>
### <primary style="font-size:24px"> Solution </primary>
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- $ \frac{y-x}{1+x y}=\frac{z-y}{1+z y} $

- $ (y-x)(1+z y)=(z-y)(1+x y) $

- $ y+z y^2-x-x y z=z+x y z-y-xy^2 $

- ($\because$ $x,y,z$ are in A.P. $\quad \quad$ $\therefore$ $2y = x + z $ or  $y-x = z - y)  $

- $ 2 x y z =y^2(x+z)=2 y^{3}$   
- $ 2 x y z =2 y^3 $   
- $ 2 y (y^2-x z)=0 $   
- $y=0 \text { or } y^2=x z $

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

### <Success style="font-size:25px"> Inverse Trigonometric Functions L-5 </Success>
### <primary style="font-size:24px"> Solution </primary>
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- For, $y =0$ ,    
- As, $ x + z=2 y = 0 \Rightarrow z = -x $

- $ \tan ^{-1} x, \tan ^{-1} (-y) =0 \quad  \tan ^{-1} = - \tan^{1} x $
- **Possibility 1 :**  $ x, 0, -x \leftarrow$ in A.P.   
- $\tan^{-1} x , 0 , \tan^{-1} (-x)$  $\leftarrow$ in A.P.

- For, $y^2 = xz$ , $x + z = 2y $  
- **Possibility 2:**  $\Rightarrow x = y = z$
  
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<footer style = "font-size: 18px">Solution $\rightarrow$ Solution $\rightarrow$ <span style = "color:blue"> Solution </span> $\rightarrow$ Problem $\rightarrow$ Solution </footer>

### <Success style="font-size:25px"> Inverse Trigonometric Functions L-5 </Success>
### <primary style="font-size:24px"> Problem </primary>
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- **<ins>Problem:5</ins>**

- If  $0 < x < 1$  then,

- $ \sqrt{1+x^2}\sqrt{[\\{x \cos(\cot ^{-1} x)+\sin(\cot ^{-1} x)\\}^2-1]}$

- is equal to ___ ?

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

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<h3 id="primary-stylefont-size24px-problem-primary-1"><primary style="font-size:24px"> Problem </primary></h3>
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<p><b><ins>Problem:6</ins></b></p>
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<p>Find the number of positive solutions satisfying the equation</p>
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<p>$\tan ^{-1}\left(\frac{1}{2 x+1}\right)+\tan ^{-1}\left(\frac{1}{4 x+1}\right)=\tan ^{-1} \frac{2}{x^2}$</p>
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<p>$x >0  $</p>
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<p>$2 x+1  >1$</p>
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<p>$ 0 < \frac{1}{2 x+1} \cdot \frac{1}{4 x+1}<1$</p>
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<footer style = "font-size: 18px">Solution $\rightarrow$ Solution $\rightarrow$ <span style = "color:blue"> Problem </span> $\rightarrow$ Solution $\rightarrow$ Solution </footer>