mathematics-class-12-unit-02-chapter-06-problems-on-trig-and-inv-trig-functions-l-1-2-fmrwxrvhxqw

### <Success style="font-size:25px"> Problems On Trig And Inv Trig Functions L-1 </Success>
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- **<ins>Problem-1</ins>**
	
- For$\quad$ $a>0, b>0$, and $c>0$ find

- $ \tan ^{-1} \sqrt{\frac{a(a+b+c)}{b c}}+\tan ^{-1} \sqrt{\frac{b(a+b+c)}{a c}} + \tan ^{-1} \sqrt{\frac{c(a+b+c)}{a b}} $

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

- $\tan ^{-1}\left[a \sqrt{\frac{a+b+c}{a b c}}\right]+\tan ^{-1}\left[b \sqrt{\frac{a+b+c}{a b c}}\right] = \tan ^{-1} X+\tan ^{-1} Y $

- where, $X=\left[a \sqrt{\frac{a+b+c}{a b c}}\right] , Y=\left[b \sqrt{\frac{a+b+c}{a b c}}\right]$

- Observe, XY = $ \frac{a + b + c}{c} > 1$

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### <Success style="font-size:25px"> Problems On Trig And Inv Trig Functions L-1 </Success>
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- **<ins>Solution</ins>** 
	
- $ \tan^{-1}\left[a\sqrt{\frac{a+b+c}{a b c}}\right]+\tan^{-1}\left[b\sqrt{\frac{a+b+c}{a b c}}\right] $

- $\because X,Y>0\quad$ & $\quad XY>1$

- $\Rightarrow \tan ^{-1} X + \tan ^{-1} Y = \pi+\tan ^{-1} \frac{X+Y}{1-X Y}$

- $=\pi+\tan ^{-1} \frac{(a+b) \sqrt{\frac{a+b+c}{a b c}}}{1-\frac{(a+b+c)}{c}} =\pi+\tan ^{-1}\left[-c \sqrt{\frac{a+b+c}{a b}}\right] $

- $ =\pi-\tan^{-1}\left[c\sqrt{\frac{a+b+c}{a b c}}\right] $$\qquad\because\tan^{-1}(-x)=-\tan ^{-1} x$

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<footer style = "font-size: 18px">Problems on trigonometric and inverse trigonometric functions lecture 6 $\rightarrow$ Problem $\rightarrow$ <span style = "color:blue"> Solution </span> $\rightarrow$ Problem $\rightarrow$ Solution </footer>

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- We rewrite the expression as follows:

- $\left(\sin ^{-1} x\right)^4+\left(\cos ^{-1} x\right)^4=\left(\sin ^{-1} x\right)^4+\left(\frac{\pi}{2}-\sin ^{-1} x\right)^4$

- Let, $\theta=\sin ^{-1} x, ~ ~$ where,   $\theta \in\left[-\frac{\pi}{2}, \frac{\pi}{2}\right] $

- Let, $ f(\theta)=\theta^4+\left(\frac{\pi}{2}-\theta\right)^4, \theta \in\left[-\frac{\pi}{2}, \frac{\pi}{2}\right] $

- Setting the derivative equal to zero, we get the point for extremum.

- $\Rightarrow f^{\prime}(\theta)=4 \theta^3-4(\pi / 2-\theta)^3=0 $

- $ \Rightarrow \theta^3-(\pi / 2-\theta)^3=0 $

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### <Success style="font-size:25px"> Problems On Trig And Inv Trig Functions L-1 </Success>
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- $\Rightarrow \theta^3=(\pi / 2-\theta)^3 $

- $\Rightarrow \theta=\frac{\pi}{2}-\theta\quad\Rightarrow  \theta=\pi / 4 $

- To determine whether this point is a maximum or minimum, we need to take second derivative,

- $f^{ \prime\prime}(\theta)=12 \theta^2+12(\frac{\pi}{2}-\theta)^2>0$

- $\therefore\quad \theta=\frac{\pi}{4}$ will give a minimum  value.

- The minimum value of $f(\theta)$ is $(\frac{\pi}{4})^4 $ + $(\frac{\pi}{2}-\frac{\pi}{4})^4= \frac {\pi^4}{ 128 }$

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### <Success style="font-size:25px"> Problems On Trig And Inv Trig Functions L-1 </Success>
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- To find the maximum value we start with the first derivative of the given function, 
	
- $f^{\prime}(\theta)=4\left[\theta^3-(\pi / 2-\theta)^3\right]$	
	
- $\rightarrow$ For,$ \frac{\pi}{4} \leqslant \theta \leqslant\frac{\pi}{2}$	
	
- $f^{\prime}(\theta) \geqslant 0 \Rightarrow f(\theta)$ is increasing in $ [{\frac{\pi}{4}},{\frac{\pi}{2}}]$

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- $\rightarrow$ For $-\frac{\pi}{2} \leqslant \theta \leqslant \frac{\pi}{4}, \quad f^{\prime}(\theta) \leqslant 0 $

- $ \Rightarrow f^{\prime}(\theta)$ is decreasing in $ [{-\frac{\pi}{2}},{\frac{\pi}{4}}]$

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- <ins>**Problem-3**</ins>

- The number of solutions to the equation$\left|\sin ^{-1}(\sin x)\right|=\cos x, \text { where, }$

- $x \in[-2 \pi, 2 \pi] \text { is }?$
	
- **<ins>Solution</ins>**

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

- $ \sin x=\sin y$

- i) $-\frac{\pi}{2} \leqslant x \leqslant \frac{\pi}{2} \quad \Rightarrow y=x$

- ii) $\frac{\pi}{2} \leqslant x \leqslant \frac{3 \pi}{2},$ then, $\quad \pi-x \in\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$

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<p>$\rightarrow$ For $\frac{3\pi}{2}\leqslant x \leqslant \frac{5\pi}{2},\quad\sin^{-1}(\sin x)=x-2\pi$</p>
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<p>$\rightarrow$ For $\frac{\pi}{2}\leqslant x \leqslant \frac{3\pi}{2},\quad\sin^{-1}(\sin x)=\pi-x$</p>
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<p>$\rightarrow$ For $-\frac{\pi}{2}\leqslant x \leqslant \frac{\pi}{2},\quad\sin^{-1}(\sin x)=x$</p>
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<p>$\rightarrow$ For $-\frac{3\pi}{2}\leqslant x \leqslant -\frac{\pi}{2},\quad\sin^{-1}(\sin x)=-(\pi+x)$</p>
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<p>$\rightarrow$ For $-\frac{5\pi}{2}\leqslant x \leqslant -\frac{3\pi}{2},\quad\sin^{-1}(\sin x)=-(x+2\pi)$</p>
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- **<ins>Solution</ins>**

- $ \sqrt{3} \sec x+\operatorname{cosec} x+2(\tan x-\cot x)=0$

- $\because\quad\sin x \neq 0\quad$,$\cos x \neq 0$

- $\frac{\sqrt{3}}{\cos x}+\frac{1}{\sin x}+2\left(\frac{\sin x}{\cos x}-\frac{\cos x}{\sin x}\right)=0$

- if $x \in $S then, $\quad\sin x \cos x \neq 0$

- $\sqrt{3} \sin x+\cos x+2\left(\sin ^2 x-\cos ^2 x\right)=0$

- $\frac{\sqrt{3}}{2} \sin x+\frac{1}{2} \cos x=\cos 2 x $

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<p>$\cos x \cos \frac{\pi}{3}+\sin x \sin \frac{\pi}{3}=\cos 2 x$</p>
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<p>$\cos \left(x-\frac{\pi}{3}\right)=\cos 2 x$</p>
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<p>$ \cos x=\cos y \Leftrightarrow x=2 n \pi \pm y $</p>
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<p>$ \cos \left(x-\frac{\pi}{3}\right)=\cos 2 x $</p>
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<p>by using identity y = $x-\frac{\pi}{3}$</p>
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<p>$ 2 x=2 n \pi \pm(x-\pi / 3)$</p>
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<p>$ 2 x=2 n \pi+(x-\pi / 3) $</p>
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<p>$ x=2 n \pi-\pi / 3$</p>
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<p>$ n=0, x=-\pi / 3 \in(-\pi, \pi), n=1, x=2 \pi-\frac{\pi}{3} \notin(-\pi, \pi)$</p>
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<p>$ x=-\pi / 3$</p>
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<p>$2 x=2 n \pi-(x-\pi / 3)$</p>
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<p>$ 3 x=2 n \pi+\pi / 3 $</p>
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<p>$ x=\frac{2 n \pi}{3}+\pi / 9$</p>
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<p>$n=0, \quad x=\pi / 9 \in(-\pi, \pi)$</p>
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<p>$n=1, \quad x=\frac{2 \pi}{3}+\pi / 9 \in(-\pi, \pi)$</p>
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<p>$n=-1, \quad x=\frac{-2 \pi}{3}+\pi / 9 \in(-\pi, n)$</p>
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<p>$x=-\frac{\pi}{3}$</p>
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