সৃজনশীলঃ ‘f(x)=cos−1⁡x…’ ময়মনসিংহ বোর্ড ২০২৫ – বিপরীত ত্রিকোনমিতিক ফাংশন ও ত্রিকোণমিতিক সমীকরণ, উচ্চতর গণিত, উচ্চতর গণিত ২য় পত্র (এইচএসসি)

দেওয়া আছে, $f\left(x\right)={\cos }^{-1}x$

এখন, $f\left(\frac{\mathrm{yz}}{\mathrm{xr}}\right)+f\left(\frac{\mathrm{zx}}{\mathrm{yr}}\right)-{\sin }^{-1}\frac{\mathrm{xy}}{\mathrm{zr}}=3\frac{\pi }{2}$

বা, ${\cos }^{-1}\frac{\mathrm{yz}}{\mathrm{xr}}+{\cos }^{-1}\frac{\mathrm{zx}}{\mathrm{yr}}-{\sin }^{-1}\frac{\mathrm{xy}}{\mathrm{zr}}=3\frac{\pi }{2}$

বা, ${\cos }^{-1}\frac{\mathrm{yz}}{\mathrm{xr}}+{\cos }^{-1}\frac{\mathrm{zx}}{\mathrm{yr}}=3\frac{\pi }{2}+{\sin }^{-1}\cdot \frac{\mathrm{xy}}{\mathrm{zr}}$

বা, ${\cos }^{-1}\{\frac{\mathrm{yz}}{\mathrm{xr}}\cdot \frac{\mathrm{zx}}{\mathrm{yr}}-\sqrt{1-{\left(\frac{\mathrm{yz}}{\mathrm{xr}}\right)}^2}\cdot \sqrt{1-{\left(\frac{\mathrm{zx}}{\mathrm{yr}}\right)}^2}\}=3\cdot \frac{\pi }{2}+{\sin }^{-1}\frac{\mathrm{xy}}{\mathrm{zz}}$

$[\because {\cos }^{-1}x+{\cos }^{-1}y={\cos }^{-1}(xy-\sqrt{1-x^2}\cdot \sqrt{1-y^2})]$

বা, $\frac{xy{z}^2}{xy{r}^2}-\sqrt{1-\frac{y^2{z}^2}{x^2{r}^2}}\sqrt{1-\frac{z^2{x}^2}{y^2{r}^2}}=\cos (3\cdot \frac{\pi }{2}+{\sin }^{-1}\frac{xy}{z})$

বা, $\frac{z^2}{r^2}-\sqrt{(1-\frac{y^2{z}^2}{x^2{r}^2})(1-\frac{z^2{x}^2}{y^2{r}^2})}=\sin {\sin }^{-1}\frac{xy}{zx}$

বা, $\frac{z^2}{r^2}-\sqrt{1-\frac{y^2{z}^2}{x^2{r}^2}-\frac{z^2{x}^2}{y^2{r}^2}+\frac{x^2{y}^2{z}^4}{x^2{y}^2{r}^4}}=\frac{xy}{zr}$

বা, $\frac{z^2}{r^2}-\frac{xy}{zr}=\sqrt{1-\frac{y^2{z}^2}{x^2{r}^2}-\frac{z^2{x}^2}{y^2{r}^2}+\frac{z^4}{r^4}}$

বা, ${(\frac{z^2}{r^2}-\frac{xy}{zr})}^2={\left(\sqrt{1-\frac{y^2{z}^2}{x^2{r}^2}-\frac{z^2{x}^2}{y^2{r}^2}+\frac{z^2}{r^3}}\right)}^2$ [বর্গ করে]

বা, ${\left(\frac{z^2}{r^2}\right)}^2-2\cdot \frac{z^2}{r^2}\cdot \frac{xy}{zr}+{\left(\frac{xy}{zr}\right)}^2=1-\frac{y^2{z}^2}{x^2{r}^2}-\frac{z^2{x}^2}{y^2{r}^2}+\frac{z^4}{r^3}$

বা, $\frac{z^4}{r^3}-2\cdot \frac{xyz}{r^3}+\frac{x^2{y}^2}{z^2{r}^2}+\frac{y^2{z}^2}{x^2{r}^2}+\frac{z^2{x}^2}{y^2{r}^2}-\frac{z^4}{r^3}=1$

বা, $\frac{x^2{y}^2}{z^2{r}^2}+\frac{y^2{z}^2}{x^2{r}^2}+\frac{z^2{x}^2}{y^2{r}^2}-\frac{2xyz}{r^3}=1$

$\therefore \frac{x^2{y}^2}{z^2{r}^2}+\frac{y^2{z}^2}{x^2{r}^2}+\frac{z^2{x}^2}{y^2{r}^2}-\frac{2xyz}{r}=1$. (প্রমাণিত)

দেওয়া আছে, $f\left(x\right)={\cos }^{-1}x$

বামপক্ষ $=f\left(\frac{1}{\sqrt{5}}\right)={\cos }^{-1}\frac{1}{\sqrt{5}}={\tan }^{-1}2$

ডানপক্ষ $={\tan }^{-1}2+\frac{1}{2}{\sin }^{-1}\frac{3}{5}-{\cot }^{-1}3$

$={\tan }^{-1}2+\frac{1}{2}{\sin }^{-1}\frac{3}{5}-{\tan }^{-1}\frac{1}{3}$

$={\tan }^{-1}2+\frac{1}{2}{\sin }^{-1}\frac{3}{5}-\frac{1}{2}\times 2{\tan }^{-1}\frac{1}{3}$

$={\tan }^{-1}2+\frac{1}{2}{\sin }^{-1}\frac{3}{5}-\frac{1}{2}{\sin }^{-1}\frac{2\times \frac{1}{3}}{1+{\left(\frac{1}{3}\right)}^2}$

$[\because 2{\tan }^{-1}x={\sin }^{-1}\frac{2x}{1+x^2}]$

$={\tan }^{-1}2+\frac{1}{2}{\sin }^{-1}\frac{3}{5}-\frac{1}{2}{\sin }^{-1}\frac{\frac{2}{3}}{1+\frac{1}{9}}$

$={\tan }^{-1}2+\frac{1}{2}{\sin }^{-1}\frac{3}{5}-\frac{1}{2}{\sin }^{-1}\frac{\frac{2}{3}}{\frac{10}{9}}$

$={\tan }^{-1}2+\frac{1}{2}{\sin }^{-1}\frac{3}{5}-\frac{1}{2}{\sin }^{-1}(\frac{2}{3}\times \frac{9}{10})$

$={\tan }^{-1}2+\frac{1}{2}{\sin }^{-1}\frac{3}{5}-\frac{1}{2}{\sin }^{-1}\frac{3}{5}$

$={\tan }^{-1}2$

$\therefore f\left(\frac{1}{\sqrt{5}}\right)={\tan }^{-1}2+\frac{1}{2}{\sin }^{-1}\frac{3}{5}-{\cot }^{-1}3.$ (দেখানো হলো)