Let y = y ( x ) be the solution of the differential equation d y d x = tan x + y sin x sec x - sin x tan x ,…

Given: $\frac{dy}{dx}=\frac{\tan x+y}{\sin {\displaystyle x}{\displaystyle \left(\sec x-\sin x\tan x\right)}}$

$\Rightarrow \frac{dy}{dx}=\frac{\tan x+y}{\sin {\displaystyle x}{\displaystyle \left(\frac{1}{\cos x}-\frac{{\sin }^{2}x}{\cos x}\right)}}$

$\Rightarrow \frac{dy}{dx}=\frac{\tan x+y}{\sin {\displaystyle x}{\displaystyle \left(\frac{{\cos }^{2}x}{\cos x}\right)}}$

$\Rightarrow \frac{dy}{dx}=\frac{\tan x+y}{\sin {\displaystyle x\cos x}}$

$\Rightarrow \frac{dy}{dx}-\frac{2y}{\sin 2x}={\sec }^{2}x$

$\Rightarrow \mathrm{IF}=e^{-2\int \mathrm{cosec}2xdx}$

$\Rightarrow \mathrm{IF}=e^{-\log \left|\mathrm{cosec}2x-\cot 2x\right|}$

$\Rightarrow \mathrm{IF}=\frac{1}{\mathrm{cosec}2x-\cot 2x}=\mathrm{cosec}2x+\cot 2x$

$\Rightarrow \mathrm{IF}=\frac{1}{\sin 2x}+\frac{\cos 2x}{\sin 2x}$

$\Rightarrow \mathrm{IF}=\frac{2{\cos }^{2}x}{2\sin x\cos x}$

$\Rightarrow \mathrm{IF}=\cot x$

$\Rightarrow y\times \cot x=\int \cot x\times {\sec }^{2}xdx$

$\Rightarrow y\times \cot x=2\int \mathrm{cosec}2xdx$

$\Rightarrow y\cot x=\log \left|\mathrm{cosec}2x-\cot 2x\right|+c$

Now, using $y\left(\frac{\pi }{4}\right)=2$ we get,

$\Rightarrow 2\cot \frac{\pi }{4}=\log \left|\mathrm{cosec}\frac{\pi }{2}-\cot \frac{\pi }{2}\right|+c$

$\Rightarrow c=2$

$\Rightarrow y\cot x=\log \left|\mathrm{cosec}2x-\cot 2x\right|+2$

Now, finding $y\left(\frac{\pi }{3}\right)$ we get,

$\Rightarrow y\cot \frac{\pi }{3}=\log \left|\mathrm{cosec}\frac{2\pi }{3}-\cot \frac{2\pi }{3}\right|+2$

$\Rightarrow y\times \frac{1}{\sqrt{3}}=\log \left|\frac{2}{\sqrt{3}}+\frac{1}{\sqrt{3}}\right|+2$

$\Rightarrow y\times \frac{1}{\sqrt{3}}=\frac{1}{2}\log 3+2$

$\Rightarrow y=\frac{\sqrt{3}}{2}\log 3+2\sqrt{3}$

$\Rightarrow y=\sqrt{3}\left(\log \sqrt{3}+2\right)$