$\begin{aligned}\cos \theta =\dfrac{12}{13}\\ \Rightarrow \tan \theta =\pm \dfrac{5}{12}\end{aligned}$

$\sin \theta =\dfrac{5}{13}$
$= \dfrac{\tan \theta +\sec \left( -\theta \right) }{\cot \theta +co\sec \left( -\theta \right) }$
$= \dfrac{\tan \theta +\sec \theta }{\cot \theta - \cosec \theta }$
$= \dfrac{\dfrac{-5}{12}-\dfrac{13}{12}}{-\dfrac{12}{5}-\dfrac{13}{5}}= \dfrac{28}{25}\times \dfrac{5}{12}$
$= \dfrac{3}{10}$

$7\sin^{2}\theta +3\cos ^{2}\theta =4$
$\Rightarrow 7\tan^{2} \theta +3 = 4\sec^{2} \theta =4 \left( 1+\tan ^{2}\theta\right)$
$\Rightarrow 7\tan^{2}\theta + 3 = 4 + 4\tan ^{2}\theta$
$\Rightarrow 3\tan ^{2}\theta =1$
$\Rightarrow \tan ^{2}\theta =\dfrac{1}{3}$
$\therefore \tan \theta =\pm \dfrac{1}{\sqrt{3}}$

$y=cot\pi=\dfrac{1}{\tan\pi}=\dfrac{1}{0}=\infty \newline সুতরাং,x=\pi\hspace{1mm}বিন্দুতে\hspace{1mm} ফাংশনটি\hspace{1mm} বিচ্ছিন্ন\hspace{1mm}।$

$\frac{2r\pi}{r} = 2\pi$

$\frac{ 2tan\theta }{ 1+tan ^ { 2}\theta }$ = $sin2\theta$ = $2sin\theta cos\theta$

$\cosec 2A - \cot 2A$
= $\frac{1}{\sin 2A}$ - $\frac{\cos 2A}{\sin 2A}$
= $\frac{1 - \cos 2A}{\sin 2A}$ - $\frac{2 \sin ^{2}A}{2 \sin A \cos A}$
= $\frac{ \sin A}{\cos A}$
= $\tan A$

$\frac{1-cos2\theta+sin2\theta}{1+cos2\theta+sin2\theta}$
= $\frac{2sin^{2}\theta+2sin\theta cos\theta}{2cos^{2}\theta+2sin\theta cos\theta}$
= $\frac{2sin\theta (sin\theta +cos\theta)}{2cos\theta (cos\theta + sin\theta)}$
= $tan\theta$

$\frac{1-cos\theta}{sin\theta}$ = $\frac{2sin^{2}\frac{\theta}{2}}{2sin\frac{\theta}{2}cos{\theta}{2}}$ = $tan\frac{\theta}{2}$

$\frac {1-\tan ^2x}{1+\tan ^2x}$ = $\cos 2x$

We know , $\sec^2\theta - \tan^2\theta$ = 1

$\Rightarrow$ $(\sec\theta + \tan\theta)$ $(\sec\theta - \tan\theta)$ = 1

$\Rightarrow$ 5 $(\sec\theta - \tan\theta)$ = 1

$\therefore$ $(\sec\theta - \tan\theta)$ = $\frac{1}{5}$

$\cos\theta$ =$\frac{1-\tan^2\frac{\theta}{2}}{1+\tan^2\frac{\theta}{2}}$ =$\frac{1-(\frac{3}{4})^2}{1+(\frac{3}{4})^2}$ = $\frac{7}{25}$

$\cos 2 \theta$ = $\frac{1 - \tan ^2\theta}{1 + \tan ^2\theta}$ = $\frac{1-t^2}{1+t^2}$

$\cos \theta$=$\frac{12}{13}$
$\tan \theta$=$\pm\frac{5}{12}$
$\tan 2 \theta$ = $\frac{2\tan\theta}{1 - \tan^2\theta}$
= $\frac{2(\pm\frac{5}{12})}{1 - \frac{25}{144}}$ = $\pm\frac{120}{119}$

$\cos 4\alpha =\cos \left( 2\times 2\alpha \right) =\cos ^{2}2\alpha-\sin ^{2}2\alpha$

$\cos \dfrac{\pi }{6}\sin \dfrac{\pi }{6} = \dfrac{1}{2}\cdot \dfrac{\sqrt{3}}{2}= \dfrac{\sqrt{3}}{4}$

$\tan ^{2}\theta =\sec ^{2}\theta -1\\ =\left( 5\right) ^{2}-1\\ =24$

$\dfrac{2\sin \dfrac{\theta }{2}\cos \dfrac{\theta }{2}}{2\cos ^{2}\dfrac{\theta }{2}} = \tan\dfrac{\theta }{2}$

$\frac{tan\theta}{1+tan^2\theta} = \frac{1}{2} \times \frac{2tan\theta}{1+tan^2\theta} = \frac{1}{2} sin2\theta= sin\theta cos\theta$

$\cos ^2 15\degree - \sin ^2 15\degree = \cos (2\times 15\degree)$
$= \cos 30 \degree = \frac {\sqrt{3}}{2}$ = $\sqrt\frac{3}{4}$

$\sin A - \sin B = \cos B - \cos A$
$\Rightarrow 2 \sin \frac{A-B}{2} \cos \frac{A+B}{2}$
$= 2sin\frac{A+B}{2} sin\frac{A-B}{2}$
$\Rightarrow tan\frac{A+B}{2} =1 = tan\frac{\pi}{4}$
$\Rightarrow \frac{A+B}{2} = \frac{\pi}{4}$
$\therefore A+B= \frac{\pi}{2}$

ত্রিকোণমিতিক MCQ গুলোর প্রায় সবগুলোই option test করে সলভ করা যায়। Option test এর ক্ষেত্রে প্রশ্নে থাকা কোণগুলোর Random কিছু value ধরে নিব। যেমনঃ প্রশ্নে বলা হয়েছে, $ABC$ হচ্ছে একটি সমকোণী ত্রিভূজ। তাই $ABC$ ত্রিভূজের যেকোনো একটি কোণ হবে $90 \degree$ ।
অর্থাৎ, $A = 90\degree, B = 30\degree , C = 60\degree$
$\cos ^2 90\degree+ \cos ^2 30\degree + \cos ^2 60\degree$
$= 0 + \frac{3}{4} + \frac{1}{4}= \frac{4}{4}=1$

ধরি, $A = 30\degree$
অর্থাৎ, $sin(30-30)+ sin(150+30)$
$= sin 0\degree + sin 180\degree =0$