সৃজনশীলঃ ‘f(a) = tan–1a,…’ রাজশাহী বোর্ড ২০২১ – বিপরীত ত্রিকোনমিতিক ফাংশন ও ত্রিকোণমিতিক সমীকরণ, উচ্চতর গণিত, উচ্চতর গণিত ২য় পত্র (এইচএসসি)

দেওয়া আছে,

$f\left(a\right)={\tan }^{-1}a,g\left(a\right)=\sin a$

$\mathrm{LHS}=2f(\sqrt{\frac{x-y}{x+y}}\tan \frac{\theta }{2})$

$=2{\tan }^{-1}(\sqrt{\frac{x-y}{x+y}}\tan \frac{\theta }{2})$

$={\sec }^{-1}\left(\frac{1+\frac{x-y}{x+y}\tan \frac{\theta }{2}}{1-\frac{x-y}{x+y}\tan \frac{\theta }{2}}\right)$

$={\sec }^{-1}\left(\frac{x+y+(x-y)\tan \frac{\theta }{2}}{x+y-(x-y)\tan \frac{\theta }{2}}\right)$

$={\sec }^{-1}\left(\frac{x(1+\tan \frac{\theta }{2})+y(1-\tan \frac{\theta }{2})}{x(1-\tan \frac{\theta }{2})+y(1+\tan \frac{\theta }{2})}\right)$

$={\sec }^{-1}(\frac{x\left(\frac{1+\tan \frac{\theta }{2}}{1+\tan \frac{\theta }{2}}\right)+y\left(\frac{1-\tan \frac{\theta }{2}}{1+\tan \frac{\theta }{2}}\right)}{x\left(\frac{1-\tan \frac{\theta }{2}}{1+\tan \frac{\theta }{2}}\right)+y\left(\frac{1+\tan \frac{\theta }{2}}{1+\tan \frac{\theta }{2}}\right)}$

$={\sec }^{-1}\left(\frac{x+y\cos \theta }{x\cos \theta +y}\right)$

$={\sec }^{-1}\left(\frac{x+y(\sin \frac{\pi }{2}-\theta )}{y+x(\sin \frac{\pi }{2}-\theta )}\right)$

$={\sec }^{-1}\frac{x+yg(\frac{\pi }{2}-\theta )}{y+xg(\frac{\pi }{2}-\theta )}$

$\text{ = RHS }$ (দেখানো হল)

দেওয়া আছে, $g\left(a\right)=\sin a$

$\therefore g(\frac{\pi }{2}-x)+g\left(x\right)=\frac{1}{\sqrt{2}}$.

$\Rightarrow \sin (\frac{\pi }{2}-x)+\sin \left(x\right)=\frac{1}{\sqrt{2}}$.

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

$\Rightarrow \frac{1}{\sqrt{2}}\cos x+\frac{1}{\sqrt{2}}\sin x=\frac{1}{2}$ [2 দ্বারা ভাগ করে]

$\Rightarrow \cos x\cdot \cos \frac{\pi }{4}+\sin x\cdot \sin \frac{\pi }{4}=\frac{1}{2}$

$\Rightarrow \cos (x-\frac{\pi }{4})=\cos \frac{\pi }{3}$

$\Rightarrow x-\frac{\pi }{4}=2n\pi \pm \frac{\pi }{3},n\in Z$

(+ve),

$\Rightarrow x-\frac{\pi }{4}=2n\pi +\frac{\pi }{3}$

$\Rightarrow x=2n\pi +\frac{\pi }{3}+\frac{\pi }{4}$

$=2n\pi +\frac{7\pi }{12},n\in Z$

$(-\mathrm{ve})$,

$\Rightarrow x-\frac{\pi }{4}=2n\pi -\frac{\pi }{3}$

$\Rightarrow x=2n\pi +\frac{\pi }{3}-\frac{\pi }{4}$

$\Rightarrow x=2n\pi -\frac{\pi }{12},n\in Z$

নির্ণেয় সমাধান: $x=2n\pi +\frac{7\pi }{12},2n\pi -\frac{\pi }{12},n\in Z$