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Question 17.2.13: Solve: sec x- tan x=√3 ....

Solve: \sec x-\tan x=\sqrt{3} .

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We have

\begin{aligned}\sec x-\tan x=\sqrt{3} & \Rightarrow \frac{1}{\cos x}-\frac{\sin x}{\cos x}=\sqrt{3} \Rightarrow(1-\sin x)=\sqrt{3} \cos x\\  \\& \Rightarrow \sqrt{3} \cos x+\sin x=1 .\end{aligned}

Thus, the given equation becomes

\sqrt{3} \cos x+\sin x=1.\qquad \qquad …(i)

Dividing both sides of (i) by \sqrt{(\sqrt{3})^{2}+1^{2}} , i.e., by 2 , we get

\begin{aligned}& \frac{\sqrt{3}}{2} \cos x+\frac{1}{2} \sin x=\frac{1}{2} \\  \\\Rightarrow & \cos x  \cos \frac{\pi}{6}+\sin x  \sin \frac{\pi}{6}=\frac{1}{2} \\  \\\Rightarrow & \cos \left(x-\frac{\pi}{6}\right)=\cos \frac{\pi}{3} \\  \\\Rightarrow & \left(x-\frac{\pi}{6}\right)=2 n \pi \pm \frac{\pi}{3}, \text { where } n \in I   [\because  \cos \theta=\cos \alpha \Rightarrow \theta=2 n \pi \pm \alpha]\\  \\\Rightarrow & \left(x-\frac{\pi}{6}\right)=\left(2 n \pi+\frac{\pi}{3}\right) \text { or }\left(x-\frac{\pi}{6}\right)=\left(2 n \pi-\frac{\pi}{3}\right) \\  \\\Rightarrow & x=2 n \pi+\left(+\frac{\pi}{3}+\frac{\pi}{6}\right) \text { or } x=2 n \pi+\left(-\frac{\pi}{3}+\frac{\pi}{6}\right), \text { where } n \in I \\  \\\Rightarrow & x=\left(2 n \pi+\frac{\pi}{2}\right) \text { or } x=\left(2 n \pi-\frac{\pi}{6}\right), \text { where } n \in I \\  \\\Rightarrow & x=\left(2 n \pi-\frac{\pi}{6}\right), \text { where } n \in I \\  \\ & \left[ \because  \sec x \text{ is not defined when  } x =(2n\pi + \frac{\pi}{2}) \right].\end{aligned}

Hence, the general solution is x=\left(2 n \pi-\frac{\pi}{6}\right) , where n \in I .

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