X = tan $\theta$, Y = cot $\theta$, Z = sin $\theta$

দেওয়া আছে, A = 30°

বামপক্ষ = tan 2A = tan(2 $\times$ 30°)

$= \tan 60° =\sqrt{3}$

ডানপক্ষ = $\frac{2\tan A}{1-ta{n}^{2}A}=\frac{2 \tan 30°}{1-{\tan }^2 30°}$

= $\frac{2 \times {\displaystyle \frac{1}{\sqrt{3}}}}{1-{\left(\frac{1}{\sqrt{3}}\right)}^2}$ [tan 30°= $\frac{1}{\sqrt{3}}$]

=$\frac{{\displaystyle \frac{2 }{\sqrt{3}}}}{1-{\displaystyle \frac{1}{3}}}$

= $\frac{{\displaystyle \frac{2 }{\sqrt{3}}}}{{\displaystyle \frac{2}{3}}}$

= $\frac{2 }{\sqrt{3}} \times \frac{3}{2}$ = $\sqrt{3}$

অর্থাৎ  tan 2A = $\frac{2\tan A}{1-ta{n}^{2}A}$(প্রমাণিত)

দেওয়া আছে, X = tan $\theta$, Y = cot $\theta$

বামপক্ষ = $\frac{X}{1-Y}+\frac{Y}{1-X}$

$= \left(\frac{\tan \theta }{1 - cot \theta }\right)+ \left(\frac{cot \theta }{1 - \tan \theta }\right)= \left(\frac{\tan \theta }{1 - {\displaystyle \frac{1}{\tan \theta }}}\right)+ \frac{{\displaystyle \frac{1}{\tan \theta }}}{1 - \tan \theta }$

$= \frac{\tan \theta }{{\displaystyle \frac{\tan \theta -1}{\tan \theta }}}+ \frac{1}{\tan \theta \left(1 - \tan \theta \right)}$

$= \tan \theta \times \frac{\tan \theta }{\tan \theta -1}+ \frac{1}{\tan \theta (1 - \tan \theta )}$

$=\frac{{\tan }^2 \theta }{\tan \theta -1}+ \frac{1}{\tan \theta (1 - \tan \theta )}$

$=\frac{{\tan }^3\theta -1}{\tan \theta \left(\tan \theta -1\right)}=\frac{\left(\tan \theta -1\right)\left(ta{n}^2\theta + tan \theta +1\right)}{\tan \theta \left(\tan \theta -1\right)}$

$=\frac{\left({\tan }^2\theta + \tan \theta +1\right)}{\tan \theta }$

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

= tan $\theta$ + 1 + cot $\theta$ = X + Y + 1 = ডানপক্ষ

$\frac{X}{1-Y}+\frac{Y}{1-X}$ = X + Y + 1 (প্রমাণিত)

দেওয়া আছে,

X = tan $\theta$ , Z = sin $\theta$

এখন, X + Z = P এবং X - Z = Q হলে

tan $\theta$ + sin $\theta$ = P _______ (i)

tan $\theta$ - sin $\theta$ = Q _______ (ii)

বামপক্ষ $=(P+Q{)}^2=(tan \theta +sin \theta +tan \theta -sin \theta {)}^2$

$=(2tan\theta {)}^2=4ta{n}^2\theta$

ডানপক্ষ = $\frac{16 PQ}{{(P-Q)}^2}$

$=\frac{16(\tan \theta +\sin \theta )(\tan \theta -\sin \theta )}{(tan\theta +sin\theta -tan\theta +sin\theta {)}^2}$

$=\frac{16({\tan }^2\theta -{\sin }^2\theta )}{(2sin\theta {)}^2}$

$=\frac{16 {\sin }^2\theta \left({\displaystyle \frac{{\tan }^2\theta }{{\sin }^2\theta }}-1\right)}{4si{n}^2\theta }$

$=4\left(\frac{{\sin }^2 \theta }{{\cos }^2 \theta }\times \frac{1}{{\sin }^2}-1\right)$

$=4 \left(\frac{1-{\cos }^2 \theta }{{\cos }^2 \theta }\right)$

$=4. \frac{{\sin }^2 \theta }{co{s}^2 \theta }$

$=4ta{n}^2 \theta$

$\therefore$ $(P+Q{)}^2=\frac{16PQ}{(P-Q{)}^2}$  (প্রমাণিত)