সৃজনশীলঃ ‘(i)ii) g(x)=2tan⁡x21+tan2⁡x2,f(x)=1−cos2⁡x21+cos2⁡x2…’ চট্টগ্রাম কলেজ ২০২৪ – বিপরীত ত্রিকোনমিতিক ফাংশন ও ত্রিকোণমিতিক সমীকরণ, উচ্চতর গণিত, উচ্চতর গণিত ২য় পত্র (এইচএসসি)

উদ্দীপক (i) হতে পাই,

$△ABC$হতে পাই, এবং$△DEF$হতে পাই,

$\cos \alpha =\frac{x}{a}\cos \beta =\frac{y}{b}$

$\therefore \alpha ={\cos }^{-1}\left(\frac{x}{a}\right)\therefore \beta ={\cos }^{-1}\left(\frac{y}{b}\right)$

দেওয়া আছে,

$\alpha +\beta =\theta$

$\Rightarrow {\cos }^{-1}\left(\frac{x}{a}\right)+{\cos }^{-1}\left(\frac{y}{b}\right)=0$

$\Rightarrow {\cos }^{-1}\{\frac{x}{a}\cdot \frac{y}{b}-\sqrt{\left\{\frac{x}{1-{\left(\frac{x}{a}\right)}^2\}\{1-{\left(\frac{y}{b}\right)}^2}\right\}}=\theta$

$\Rightarrow \frac{xy}{ab}-\sqrt{(1-\frac{x^2}{a^2})(1-\frac{y^2}{b^2})}=\cos \theta$

$\Rightarrow \frac{xy}{ab}-\cos \theta =\sqrt{(1-\frac{x^2}{a^2})(1-\frac{y^2}{b^2})}$

$\Rightarrow {(\frac{xy}{ab}-\cos \theta )}^2={\left\{\sqrt{(1-\frac{x^2}{a^2})(1-\frac{y^2}{b^2})}\right\}}^2$ [বর্গ করে]

$\Rightarrow \frac{x^2{y}^2}{a^2{b}^2}-\frac{2xy\cos \theta }{ab}+{\cos }^2\theta =1-\frac{x^2{y}^2}{a^2{b}^2}-\frac{x^2}{a^2}-\frac{y^2}{b^2}$

$\Rightarrow -\frac{2xy\cos \theta }{ab}+{\cos }^2\theta -1=-\frac{x^2}{a^2}-\frac{y^2}{b^2}$

$\Rightarrow -(\frac{2xy\cos \theta }{ab}+1-{\cos }^2\theta )=-(\frac{x^2}{a^2}+\frac{y^2}{b^2})$

$\therefore {\sin }^2\theta +\frac{2xy\cos \theta }{ab}=\frac{x^2}{a^2}+\frac{y^2}{b^2}$ (প্রমাণিত)

উদ্দীপক (ii) অনুসারে পাই,

$g\left(x\right)=\frac{2\tan x/2}{1+{\tan }^{2}x/2}$

$\therefore g\left(x\right)=\sin x$

এবং,

$f\left(x\right)=\frac{1-{\tan }^2\frac{x}{2}}{1+{\tan }^{2}x/2}$

$\therefore f\left(x\right)=\cos x$

$4g(\pi /2-\theta )f\left(2\theta \right)+f\left(3\theta \right)-1=0$

$\Rightarrow 4\sin (\pi /2-\theta )\cos 2\theta \cos 3\theta \theta -1=0$

$\Rightarrow 4\cos \theta \cos 2\theta \cos 3\theta -1=0$

= $2\cos 2\theta (2\cos 3\theta \cos \theta )\theta -1=0$

= $2\cos 2\theta (\cos 4\theta +\cos 2\theta )-1=0$

= $2\cos 4\theta \cos 2\theta +2{\cos }^{2}2\theta -1=0$

= $2\cos 4\theta \cos 2\theta +\cos 4\theta =0$

$\Rightarrow \cos 4\theta (2\cos 2\theta +1)=0$

হয়, $\cos 4\theta =0$

$\Rightarrow u\theta =(2n+1)\pi /2,n\in z$

$\therefore \theta =(2n+1)\frac{\pi }{8},n\in z$

আবার,

$2\cos 2\theta +1=0$

$\Rightarrow \cos 2\theta =-1/2$

$\Rightarrow \cos 2\theta =\cos \frac{2\pi }{3}$

$\Rightarrow 2\theta =2n\pi \pm \frac{2\pi }{3}.$

$\therefore \theta =n\pi \pm \frac{\pi }{3};n\in z$

$\therefore n=0$ হলে, $\theta =\pi /8,-\pi /3,\pi /3$

$n=-1$ হলে, $\theta =-\pi /8,-\frac{2\pi }{3}$

$n=-2$ হলে, $\theta =-3\pi /8$

$n=-3$ হলে, $\theta =-5x/8$

$n=-4$ হলে, $\theta =-7\pi /8$

$n=1$ হলে, $\theta =\frac{3\pi }{8},\frac{2\pi }{3}$

$n=2$ হলে, $\theta =\frac{5\pi }{8}$

$n=3$ হলে, $\theta =\frac{7\pi }{8}$

$\therefore$ নির্ণেয় সমাধান: $-\frac{7\pi }{8},\frac{-5\pi }{8},-\frac{3\pi }{8},-\frac{\pi }{3},\frac{-\pi }{8}-2\pi /3,\frac{\pi }{8},\pi /3,\frac{3\pi }{8},\frac{5\pi }{8},\frac{2\pi }{3},\frac{7\pi }{8}$