HSC - উচ্চতর গণিত ২য় পত্র→অধ্যায়-০৭ঃ বিপরীত ত্রিকোণমিতিক ফাংশন ও ত্রিকোণমিতিক সমীকরণ→All Topics

$\boxed{\mathbf{~ক~}}$বামপক্ষ $=\operatorname{cosec} \sin ^{-1} \tan \sec ^{-1} \frac{x}{y}$$\begin{array}{l}\Rightarrow \operatorname{cosec} \sin ^{-1} \tan \tan ^{-1} \frac{\sqrt{x^{2}-y^{2}}}{y^{0}} \\\Rightarrow \operatorname{cosec} \sin ^{-1} \frac{\sqrt{x^{2}-y^{2}}}{y} \\\Rightarrow \operatorname{cosec} \operatorname{cosec}^{-1} \frac{y}{\sqrt{x^{2}-y^{2}}} \\\Rightarrow \frac{y}{\sqrt{x^{2}-y^{2}}}\end{array}$= ডানপক্ষ [প্রমাণিত]$\\$$\boxed{\mathbf{~খ~}}$দেওয়া আছে, $f(x)=\operatorname{cosec} x-\cot x$এবং $f(\theta)=\frac{3}{4}$$\therefore \operatorname{cosec} \theta-\cot \theta=\frac{3}{4}$$\begin{array}{l}\Rightarrow \frac{1}{\sin \theta}-\frac{\cos \theta}{\sin \theta}=\frac{3}{4} \\\Rightarrow \frac{1-\cos \theta}{\sin \theta}=\frac{3}{4} \\\Rightarrow 4-4 \cos \theta=3 \sin \theta \\\Rightarrow 3 \sin \theta-4=-4 \cos \theta \\\Rightarrow (3 \sin \theta-4)^{2}=(-4 \cos \theta)^{2} \\\Rightarrow 9 \sin ^{2} \theta-24 \sin \theta+16=16 \cos ^{2} \theta \\\Rightarrow 9 \sin ^{2} \theta-24 \sin \theta+16=16\left(1-\sin ^{2} \theta\right) \\\Rightarrow 9 \sin ^{2} \theta-24 \sin \theta+16=16-16 \sin ^{2} \theta \\\Rightarrow 25 \sin ^{2} \theta-24 \sin \theta=0 \\\Rightarrow 25 \sin ^{2} \theta=24 \sin \theta \\\Rightarrow \sin \theta=\frac{24}{25}\end{array}$$\therefore \theta=\sin ^{-1} \frac{24}{25}$[প্রমাণিত]$\\$$\boxed{\mathbf{~গ~}}$ দেওয়া আছে, $\begin{array}{l}g(5 \theta)-\sqrt{3} g(\theta)=g(3 \theta)\\\Rightarrow \sin 5 \theta-\sqrt{3} \sin \theta=\sin 3 \theta\\\Rightarrow \sin 5 \theta-\sin 3 \theta=\sqrt{3} \sin \theta\\\Rightarrow 2 \cos \left(\frac{5 \theta+3 \theta}{2}\right) \sin \left(\frac{5 \theta-3 \theta}{2}\right)=\sqrt{3} \sin \theta\\\Rightarrow 2 \cos 4 \theta \sin \theta=\sqrt{3} \sin \theta\\\Rightarrow 2 \cos 4 \theta \sin \theta-\sqrt{3} \sin \theta=0\\\Rightarrow \sin \theta(2 \cos 4 \theta-\sqrt{3})=0\end{array}$হয়, $, \sin \theta=0$$\therefore \theta=\mathrm{n} \pi ; \mathrm{n} \in \mathbb{Z}$ অথবা, $2 \cos 4 \theta-\sqrt{3}=0$$\begin{array}{l}\Rightarrow 2 \cos 4 \theta=\sqrt{3}\\\Rightarrow \cos 4 \theta=\frac{\sqrt{3}}{2}\\\Rightarrow \cos 4 \theta=\cos \frac{\pi}{6}\\\Rightarrow 4 \theta=2 n \pi \pm \frac{\pi}{6} ; n \in \mathbb{Z}\\\therefore \theta=\frac{1}{4}\left(2 \mathrm{n} \pi \pm \frac{\pi}{6}\right) ; \mathrm{n} \in \mathbb{Z}\end{array}$$\therefore$ নির্ণেয় সমাধান $\theta=n \pi, \frac{1}{4}\left(2 n \pi \pm \frac{\pi}{6}\right) ; n \in \mathbb{Z}$