সৃজনশীলঃ ‘f(x)=cosecx;g(x)=tan⁡x…’ ২০২৫ – বিপরীত ত্রিকোনমিতিক ফাংশন ও ত্রিকোণমিতিক সমীকরণ, উচ্চতর গণিত, উচ্চতর গণিত ২য় পত্র (এইচএসসি)

খ দেওয়া আছে, $f\left(x\right)=\mathrm{cosec}x$
এখানে , $2{\tan }^{-1}\left\{f\right(x\left)\right\}={\cot }^{-1}\left(\frac{\cos x}{2}\right)$
বা, $2{\tan }^{-1}\left(\mathrm{cosec}x\right)={\cot }^{-1}\left(\frac{\cos x}{2}\right)$
বা, ${\tan }^{-1}\frac{2\mathrm{cosec}x}{1-{\mathrm{cosec}}^{2}x}={\tan }^{-1}\left(\frac{2}{\cos x}\right)$
বা, $\frac{2\mathrm{cosec}x}{1-{\mathrm{cosec}}^{2}x}=\frac{2}{\cos x}$
বা, $\frac{\mathrm{cosec}x}{-{\cot }^{2}x}=\frac{1}{\cos x}$
বা, $-{\tan }^{2}x\cdot \mathrm{cosec}x=\frac{1}{\cos x}$
বা, $-{\tan }^{2}x\cdot \frac{1}{\sin x}=\frac{1}{\cos x}$
বা, $-{\tan }^{2}x=\frac{\sin x}{\cos x}$ 

বা, $-{\tan }^{2}x=\tan x$
বা, ${\tan }^{2}x+\tan x=0$
বा, $\tan x(1+\tan x)=0$ 

$\begin{array}{cc}\therefore & \tan x=0\\ \therefore & x=n\pi ;n\in Z\end{array}$ 

অথবা, $1+\tan x=0$
বা, $\tan x=-1=-\tan \frac{\pi }{4}$
বা, $\tan x=\tan (-\frac{\pi }{4})$ 

$\therefore x=n\pi +(-\frac{\pi }{4})=n\pi -\frac{\pi }{4},n\in Z$ 

নির্ণেয় সমাধান : $x=n\pi ,n\pi -\frac{\pi }{4},n\in Z$. 

দেওয়া আছে, $f\left(x\right)=\mathrm{cosec}x$ 

$g\left(x\right)=\tan x$ 

বামপক্ষ $={\tan }^{-1}\left\{f({\cos }^{-1}x)\right\}-{\tan }^{-1}\left\{g({\sin }^{-1}x)\right\}$ 

$\begin{array}{cc}& ={\tan }^{-1}\left\{\mathrm{cosec}({\cos }^{-1}x)\right\}-{\tan }^{-1}\{\tan ({\sin }^{-1}x)\}\\ & ={\tan }^{-1}\left\{\mathrm{cosec}\left({\mathrm{cosec}}^{-1}\frac{1}{\sqrt{1-x^2}}\right)\right\}-{\sin }^{-1}x\end{array}$ 

$\begin{array}{cc}& ={\tan }^{-1}\frac{1}{\sqrt{1-x^2}}-{\sin }^{-1}x\\ & ={\tan }^{-1}\frac{1}{\sqrt{1-x^2}}-{\tan }^{-1}\frac{x}{\sqrt{1-x^2}}\\ & ={\tan }^{-1}\frac{\frac{1}{\sqrt{1-x^2}}-\frac{x}{\sqrt{1-x^2}}}{1+\frac{1}{\sqrt{1-x^2}}\times \frac{x}{\sqrt{1-x^2}}}\\ & ={\tan }^{-1}\frac{\frac{1-x}{\sqrt{1-x^2}}}{\frac{1-x^2+x}{1-x^2}}\end{array}$ 

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

$\therefore {\tan }^{-1}\left\{f({\cos }^{-1}x)\right\}-{\tan }^{-1}\{g({\sin }^{-1}x\}$$={\tan }^{-1}\frac{(1-x)\sqrt{1-x^2}}{1+x-x^2}$ (দেখানো হল )