Differentiation (Part-38) | S N De

Problem 1(vi)

\[ \begin{aligned} 3\sin^2\theta+7\cos^2\theta &= 6 \\[4pt] \Rightarrow\quad 3\tan^2\theta+7 &= 6\sec^2\theta \qquad [\text{Dividing both sides by }\cos^2\theta] \\[4pt] \Rightarrow\quad 3\tan^2\theta+7 &= 6(1+\tan^2\theta) \\[4pt] \Rightarrow\quad 6+6\tan^2\theta-3\tan^2\theta-7 &= 0 \\[4pt] \Rightarrow\quad 3\tan^2\theta &= 1 \\[4pt] \Rightarrow\quad \tan^2\theta &= \frac{1}{3} \\[4pt] \Rightarrow\quad \tan\theta &= \pm\frac{1}{\sqrt3} \\[4pt] \Rightarrow\quad \tan\theta &= \tan\left(\pm\frac{\pi}{6}\right) \\[4pt] \therefore\quad \theta &= n\pi\pm\frac{\pi}{6}, \qquad n\in\mathbb Z \end{aligned} \]

Problem 1(vii)

\[ \begin{aligned} \tan p\theta &= \cot q\theta \\[4pt] \Rightarrow\quad \tan p\theta &= \tan\left(\frac{\pi}{2}-q\theta\right) \\[4pt] \Rightarrow\quad p\theta &= n\pi+\left(\frac{\pi}{2}-q\theta\right) \\[4pt] \Rightarrow\quad (p+q)\theta &= \frac{(2n+1)\pi}{2} \\[4pt] \Rightarrow\quad \theta &= \frac{(2n+1)\pi}{2(p+q)}, \qquad n\in\mathbb Z \end{aligned} \]

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Problem 1(viii)

\[ \begin{aligned} \tan5\theta+\cot2\theta &= 0 \\[4pt] \Rightarrow\quad \frac{\sin5\theta}{\cos5\theta} + \frac{\cos2\theta}{\sin2\theta} &=0 \\[4pt] \Rightarrow\quad \frac{ \sin5\theta\sin2\theta+ \cos5\theta\cos2\theta }{ \cos5\theta\sin2\theta } &=0 \\[4pt] \Rightarrow\quad \cos5\theta\cos2\theta+ \sin5\theta\sin2\theta &=0 \\[4pt] \Rightarrow\quad \cos(5\theta-2\theta)&=0 \\[4pt] \Rightarrow\quad \cos3\theta&=0 \\[4pt] \Rightarrow\quad 3\theta&=\frac{(2n+1)\pi}{2} \\[4pt] \Rightarrow\quad \theta&=\frac{(2n+1)\pi}{6}, \qquad n\in\mathbb Z \end{aligned} \]

Problem 1(ix)

\[ \begin{aligned} \sin5\theta+\sin2\theta &=0 \\[4pt] \Rightarrow\quad 2\sin\frac{5\theta+2\theta}{2} \cos\frac{5\theta-2\theta}{2} &=0 \\[4pt] \Rightarrow\quad 2\sin\frac{7\theta}{2} \cos\frac{3\theta}{2} &=0 \end{aligned} \]

Therefore,

\[ \sin\frac{7\theta}{2}=0 \qquad\text{or}\qquad \cos\frac{3\theta}{2}=0. \]

Hence, from the first equation:

\[ \frac{7\theta}{2}=n\pi \quad\Rightarrow\quad \theta=\frac{2n\pi}{7}. \]

Again, from the second equation:

\[ \frac{3\theta}{2} = \frac{(2n+1)\pi}{2} \quad\Rightarrow\quad \theta= \frac{(2n+1)\pi}{3}, \qquad n\in\mathbb Z. \]

Problem 1(x)

\[ \begin{aligned} \tan^2x &= 3\csc^2x-1 \\[4pt] \Rightarrow\quad 1+\tan^2x &= 3\csc^2x \\[4pt] \Rightarrow\quad \sec^2x &= 3\csc^2x \\[4pt] \Rightarrow\quad \frac{1}{\cos^2x} &= \frac{3}{\sin^2x} \\[4pt] \Rightarrow\quad \sin^2x &= 3\cos^2x \\[4pt] \Rightarrow\quad \tan^2x &= 3 \\[4pt] \Rightarrow\quad \tan x &= \pm\sqrt3 \\[4pt] \Rightarrow\quad \tan x &= \tan\left(\pm\frac{\pi}{3}\right) \\[4pt] \Rightarrow\quad x &= n\pi\pm\frac{\pi}{3}, \qquad n\in\mathbb Z \end{aligned} \]

Problem 1(xi)

\[ \begin{aligned} \tan x+\cot x &=2 \\[4pt] \Rightarrow\quad p+\frac{1}{p} &=2, \qquad p=\tan x \\[4pt] \Rightarrow\quad p^2+1 &=2p \\[4pt] \Rightarrow\quad p^2-2p+1 &=0 \\[4pt] \Rightarrow\quad (p-1)^2 &=0 \\[4pt] \Rightarrow\quad p&=1 \\[4pt] \Rightarrow\quad \tan x &= 1 = \tan\frac{\pi}{4} \\[4pt] \Rightarrow\quad x&= n\pi+\frac{\pi}{4}, \qquad n\in\mathbb Z \end{aligned} \]

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