mathematics-class-12-unit-02-chapter-07-problems-on-trig-and-inv-trig-functions-l-2-2-posdd6yvo9y

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<p><ins><strong>Problem-1</strong></ins></p>
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<p>Let, $\quad\frac{-\pi}{6}<\theta<\frac{-\pi}{12}$, suppose $\alpha$ and $\beta$ are the roots of the equation $x^2-2 x \sec \theta+1=0$ , and $\alpha_2$ and $\beta_2$ are the root of the equation $x^2+2 x \tan \theta-1=0.\quad \text{If} \quad \alpha_ 1>\beta_ 1$, and $\alpha_ 2>\beta_ 2$ then, $\alpha_ 1+\beta_2$ equals?</p>
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<p><ins><strong>Solution</strong></ins></p>
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<p>$ x^2-2 x \sec \theta+1=0$</p>
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<p>$ \left(\alpha_1, \beta_1\right)  =\frac{2 \sec \theta \pm \sqrt{4 \sec ^2 \theta-4}}{2}$</p>
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<p>$ =\sec \theta \pm \sqrt{\sec ^2 \theta-1}$</p>
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- $ \sec ^2 \theta  =1+\tan ^2 \theta$

- $ \left(\alpha_1, \beta_1\right)=\sec \theta \pm \tan \theta$

- $ =\left(\frac{1 \pm \sin \theta}{\cos \theta}\right)$

- $ \theta \in\left(-\frac{\pi}{6},-\frac{\pi}{12}\right) \Rightarrow \quad\sin \theta<0, \cos \theta>0$

- $\frac{\alpha_1=(1-\sin \theta)}{\cos \theta}$

- $ x^2+2 x \tan \theta-1=0$

- $ \left(\alpha_2, \beta_2\right)  =\frac{-2 \tan \theta \pm \sqrt{4 \tan ^2 \theta+4}}{2}$

- $ =-\tan \theta \pm \sqrt{1+\tan^2 \theta } =  \sec^{2}\theta$    $\quad$ $ [\because 1+ \tan^2 \theta = {\sec ^2 \theta}]$

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- $ \begin{aligned}
& \left(\alpha_ 2, \beta_2\right)=-\tan \theta \pm \sec \theta \\\\
&=\frac{ \pm 1-\sin \theta}{\cos \theta} \\\\
&\theta \in\left(\frac{-\pi}{6}, \frac{-\pi}{12}\right)\Rightarrow \sin \theta<0, \cos \theta>0 \\\\
&\frac{1-\sin \theta}{\cos \theta}>\frac{-1-\sin \theta}{\cos \theta}=\beta_2 \\\\
&\alpha_1=\frac{1-\sin \theta}{\cos \theta} \\\\
&\alpha_1+\beta_2=-2 \tan \theta \end{aligned} $

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- <ins>**Problem-2**</ins>

- The number of possible values of $\theta$ where $0<\theta<\pi$, for which the system of equation

- $ (y+z) \cos 3 \theta=  (x y z) \sin 3 \theta$

- $ x  \sin 3 \theta=  \frac{2\cos 3 \theta}{y} +\frac{2 \sin 3 \theta}{z}$

- $ x y z \sin 3 \theta =(y+2 z) \cos 3 \theta +y \sin 3 \theta$

- have a solution $(x_0,y_0,z_0)$ with $\quad y_0 z_0\neq 0$ is ?

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<p><strong><ins>Solution</ins></strong></p>
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<p>$ (y+z) \cos 3 \theta=  (xyz) \sin 3 \theta$</p>
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<p>$ xyz \sin 3 \theta  =2 z \cos 3 \theta + 2y \sin 3 \theta$</p>
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<p>$ xyz \sin 3 \theta  = (y+2z)\cos 3 \theta + 2y \sin 3 \theta$</p>
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<p>$(y+3) \cos 3 \theta=  2 z \cos 3\theta+2 y \sin 3 \theta$</p>
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<p>$(y+3) \cos 3 \theta=  (y+2 z) \cos 3 \theta+y \sin 3 \theta$</p>
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<p>$=  (y\cos 3 \theta)+2 z \cos 3 \theta +y \sin 3 \theta$</p>
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<p>$ {2 z \cos 3} \theta+2 y \sin 3 \theta =y \cos 3 \theta+2 {z\cos 3} \theta+y \sin 3 \theta$</p>
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<p>$ y \sin 3 \theta=y \cos 3 \theta$</p>
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<p>$ y(\sin 3 \theta-\cos 3 \theta)=0$</p>
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<p>$ \rightarrow \sin 3 \theta-\cos 3 \theta=0, \quad y \neq 0$</p>
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<p>$ \rightarrow(y+z) \cos 3 \theta =(2 z+2 y) \cos 3 \theta$</p>
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<p>$ (y+z)=0 \quad \quad(y+z) \cos 3 \theta=0$</p>
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<p>$ \sin 3 \theta-\cos 3 \theta=0$</p>
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<p>$ \frac{1}{\sqrt{2}} \sin 3 \theta-\frac{1}{\sqrt{2}} \cos 3 \theta=0$</p>
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<p>$ \sin 3 \theta \cos \frac{\pi}{4}-\sin \frac{\pi}{4} \cos 3 \theta=0$</p>
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<p>$ \sin (3 \theta-\frac{\pi}{4})=0$</p>
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<p>$ 3 \theta-\frac{\pi}{4}=n \pi \quad$ for some integer n</p>
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<p>$ \theta=\frac{n \pi}{3}+\frac{\pi}{12} \in(0, \pi)$</p>
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<p>$ \theta=\left(\frac{\pi}{12}, \frac{5 \pi}{12}, \frac{3 \pi}{4}\right)$</p>
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- <ins>**Problem-3**</ins>

- The number of distinct solutions of the equation

- $\frac{5}{4} \cos ^2 2 x+\cos ^4 x+\sin ^4 x+\cos ^6 x + \sin ^6 x=2$

- $ \text{in the interval} [0,2 \pi] $ is?

- <ins>Solution</ins>

- $ \left(\sin ^2 x+\cos ^2 x\right)^3=1$

- $[\because (A+B)^3= A^3+B^3+3AB^2+3A^2B]$

- $ \sin ^6 x+\cos ^6 x +3 \sin ^2 x \cos ^4 x +3 \cos ^2 x \sin ^4 x=1$

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<p>$ \sin ^6 x+\cos ^6 x=1 - 3 \sin ^2 x \cos ^4 x -3 \cos^2 x \sin^4 x$</p>
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<p>$\rightarrow  \sin ^6 x+\cos ^6 x=1-3 \sin ^2 x \cos ^4 x -3\cos^2 x \sin^4 x$</p>
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<p>$\left(\sin ^2 x+\cos ^2 x\right)^2=1$</p>
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<p>$ \sin ^4 x+\cos ^4 x +2 \sin^2 x \cos^2 x =1$</p>
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<p>$ \rightarrow  \sin ^4 x+\cos ^4 x=1 - 2 \sin ^2 x \cos ^2 x$</p>
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<p>$ \cos 2 x=\left(\cos ^2 x - \sin ^2 x \right)$</p>
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<p>$ \rightarrow \cos^2 2 x = \cos^4 x+\sin^4 x -2 \sin^2 x \cos^2 x$</p>
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<p>$ \frac{5}{4} \cos ^2 2x  +\left(1-2 \sin ^2 x \cos ^2 x\right)
+1-3 \sin ^2 x \cos ^4 x $</p>
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<p>$-3 \cos^2 x \sin^4 x = 2$</p>
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<p>$ \frac{5}{4} \cos ^2 2 x=  2 \sin ^2 x \cos ^2 x+3 \sin ^2 x \cos ^2 x (\sin ^2 x + \cos^2 x)$</p>
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<p>$ \frac{5}{4} \cos ^2 x =5 \sin ^2 x \cos ^2 x$</p>
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<p>$ \cos ^2 2 x=  4 \sin ^2 x \cos ^2 x =  \left(2 \sin x \cos x\right)^2$</p>
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<p>$ \cos ^2 2x  =\sin ^2 2 x$</p>
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<p>$ \cos ^2 2 x=\sin ^2 2 x$</p>
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<p><ins><strong>Problem-4</strong></ins></p>
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<p>Prove that the values of the function $\left(\frac{\sin x \cos 3 x}{\sin 3 x \cos x}\right) $ do not lie between $\frac{1}{3}$ and 3 for any real $x$.</p>
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<p><strong><ins>Solution</ins></strong></p>
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<p>$ \frac{\tan x}{\tan 3 x}$</p>
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<p>$\tan 3 x =\frac{3 \tan x-\tan ^3 x}{1-3 \tan ^2 x}$</p>
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<p>$ =\frac{1-3 \tan ^2 x}{3-\tan ^2 x}  =\frac{1-3 a}{3-a}$</p>
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<p>$ a = \tan ^2 x \geqslant 0 \quad \quad a \in[0, \infty]$</p>
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<p>$\frac{1-3 a}{3-a} \notin\left(\frac{1}{3}, 3\right) \quad a \in[0, \infty]$</p>
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<p>$\frac{1-3 a}{3-a}=\frac{(9-3 a)-8}{3-a}=3-\frac{8}{3-a}=3+\frac{8}{a-3}$</p>
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<p>i) $ a \in[0,3]$</p>
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<p>$ 3 \geqslant 3-a>0$</p>
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<p>$ -3 \leqslant a-3 < 0$</p>
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<p>$ -\infty<\frac{8}{a-3} \leqslant \frac{-8}{3}$</p>
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- <ins>**Problem-5**</ins>

- $\sum_{k=1}^{13} \frac{2}{2 \sin \left(\frac{\pi}{4}+\frac{(k-1) \pi}{6}\right) 2 \sin \left(\frac{\pi}{4}+\frac{k \pi}{6}\right)}$

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

- $ 2 \sin A \sin B=\cos (A-B)-\cos (A+B)$

- $ =\cos \frac{\pi}{6}-\cos \left(\frac{\pi}{2}+\frac{(2 k-1) \pi}{6}\right)$

- $ =\cos \frac{\pi}{6}+\frac{\sin (2 k-1) \pi}{6} \quad [\because \cos (\frac{\pi}{2}+\theta) \\  =-\sin \theta]$

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- $ \sum_{k=1}^{13} \frac{2}{[\frac{\sqrt{3}{}}2 +\sin \frac{(2 k-1) \pi}{6}]}$

- $ (k+6)^{th}$term: $ \sin (\frac{2(k+6)-1) \pi}{6}$

- $ =\sin \left[\frac{(2 k-1) \pi}{6}+\frac{12 \pi}{6}\right]$

- $ =\sin \left[\frac{(2 k-1) \pi}{6}+2 \pi\right]$

- $ =\sin \frac{(2 k-1) \pi}{6} \qquad [\sin \frac{25 \pi}{6}]$

- $ =\sin (4 \pi + \frac{\pi}{6})=\sin (\frac{\pi}{6})$

- $ =k^{th}$ term     
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- $ \left(\sum_{k=1}^{6} \frac{4}{[\frac{\sqrt{3}{}}2 +\sin \frac{(2 k-1) \pi}{6}]}\right) + \frac{2}{[\frac{\sqrt{3}{}}2 +\sin \frac{\pi}{6}]}$
- $=\frac{2}{\left[\frac{\sqrt{3}}{2}+\frac{1}{2}\right]}=\frac{4}{[\sqrt{3}+1]}$

- $ =\frac{24 (\sqrt{3}-1)}{2}=2(\sqrt{3}-1)$

- $ \frac{4}{\left[\frac{\sqrt{3}}{2}+\sin \frac{(2 k-1) \pi}{6}\right]}$

- $ k=1 \quad \frac{4}{\left[\frac{\sqrt{3}}{2}+\frac{1}{2}\right]}$

- $ k=2 \quad \frac{4}{\left[\frac{\sqrt{3}}{2}+1\right]}$

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<p>$ k=3 \quad \frac{4}{\left[\frac{\sqrt{3}}{2}+\frac{1}{2}\right]}$ $\quad$ $[\because \sin \frac{5 \pi}{6} =\sin \frac{\pi}{6} =\frac{1}2]$</p>
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<p>$ k=4 \quad \frac{4}{\left[\frac{\sqrt{3}}{2}-\frac{1}{2}\right]}$</p>
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<p>$ k=5 \quad \frac{4}{\left[\frac{\sqrt{3}}{2}-1\right]} $</p>
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<p>$ k=6 \quad \frac{4}{\left[\frac{\sqrt{3}}{2}-\frac{1}{2}\right]}$  $\quad$ $[\because \sin \frac{11 \pi}{6} =-\sin \frac{\pi}{6}]$</p>
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