$AB=\left[\begin{array}{ccc}3x& -4x& 2x\\ -2x& x& 0\\ -x& -x& x\end{array}\right]\left[\begin{array}{ccc}x& 2x& -2x\\ 2x& 5x& -4x\\ 3x& 7x& -5x\end{array}\right]$

$=x\left[\begin{array}{ccc}3& -4& 2\\ -2& 1& 0\\ -1& -1& 1\end{array}\right] .x \left[\begin{array}{ccc}1& 2& -2\\ 2& 5& -4\\ 3& 7& -5\end{array}\right]$$=x^2\left[\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 0& 0& 1\end{array}\right] =\left[\begin{array}{ccc}{x}^2& 0& 0\\ 0& x^2& 0\\ 0& 0& x^2\end{array}\right]$

অনুরূপভাবে, $BA=\left[\begin{array}{ccc}{x}^2& 0& 0\\ 0& x^2& 0\\ 0& 0& x^2\end{array}\right]$

অতএব, AB = BA

অতএব,  $x^{2}I\Rightarrow \frac{1}{x^2}A =B^{-1} I \Rightarrow B^{-1}=B^{-1}=\frac{A}{x^2}=\left[\begin{array}{ccc}\frac{3}{x}& -\frac{4}{x}& \frac{2}{x}\\ -\frac{2}{x}& \frac{1}{x}& 0\\ -\frac{1}{x}& -\frac{1}{x}& \frac{1}{x}\end{array}\right]$

অতএব, $B^{-1}=\frac{A}{x^2}$

$\frac{k}{\sqrt{se{c}^2\theta +\cos e{c}^2\theta }}=2 \Rightarrow \frac{k^2}{{\displaystyle \frac{1}{{\cos }^2\theta }}+{\displaystyle \frac{1}{{\sin }^2\theta }}}=4$

$\Rightarrow \frac{k^2{\sin }^2{\cos }^2\theta }{{\sin }^2\theta +{\cos }^2\theta }=4$

$\Rightarrow 4k^2{\sin }^2\theta {\cos }^2\theta = 16 \Rightarrow {\left(k\sin 2\theta \right)}^2=4^{2 }... ... ... \left(i\right)$

$\left|\frac{-k\cos 2\theta }{\sqrt{{\cos }^2\theta +{\sin }^2\theta }}\right|=3\Rightarrow {\left(k\cos 2\theta \right)}^2=3^2... ... ... \left(ii\right)$

(i) + (ii) করে পাই, $k^2({\sin }^{2}2\theta +{\cos }^{2}2\theta )=25 \Rightarrow k =\pm 5$

$a \tan \theta +b sec\theta =c \Rightarrow a \tan \theta -c =-bsec\theta$

$a^2 {\tan }^2\theta + c^2 -2ca \tan \theta = b^2 + b^2 {\tan }^2\theta$

$\Rightarrow \left(a^2-b^2\right) {\tan }^2\theta - 2ca \tan \theta + c^2-b^2=0$

$\tan \alpha + \tan \beta =\frac{2ca}{a^2-b^2}; \tan \alpha \tan \beta =\frac{c^2-b^2}{a^2-b^2}$

$L.H.S= \tan (\alpha +\beta )=\frac{\tan \alpha + \tan \beta }{1- \tan \alpha \tan \beta }= \frac{{\displaystyle \frac{2ca}{a^2-b^2}}}{1-{\displaystyle \frac{c^2-b^2}{a^2-b^2}}}$

$=\frac{2ca}{a^2-b^2-c^2+b^2}= \frac{2ca}{a^2-c^2}=R.H.S \left(Proved\right)$

$\tan (In y) = x \Rightarrow In y = {\tan }^{-1}x \Rightarrow y = e^{\tan -1}x$

$y_1=\frac{e^{\tan -1}x}{1+x^2}\Rightarrow y^2=\frac{1}{{(1+x^2)}^2}{e}^{\tan -1}x + e^{\tan -1} \frac{(-2x)}{{(1+x^2)}^2}$

$at x = 0, y_2\left(0\right) =\frac{1}{1}\times e^0+e^0\times \frac{0}{1}=1$

অতএব, y₂(0) = 1

$\underset{x\to 0}{\lim }\frac{e^x+e^{-x}-2}{x}\left[\frac{0}{0}form\right]$

$\underset{x\to 0}{\lim }\frac{e^x-e^{-x}}{1}[L\text{'}Hospital Rule] = 0$

ধরি, $x = 2 \sin \theta \Rightarrow dx =2 \cos \theta d\theta$

x = 1 হলে, $\theta = \frac{\pi }{6} ; x =-1$হলে, $\theta =-\frac{\pi }{6}$

${\int }_{-\frac{\pi }{6}}^{\frac{\pi }{6}} 4 {\sin }^2\theta \times 2 \cos \theta \times 2 \cos \theta d\theta$

$={\int }_{-\frac{\pi }{6}}^{\frac{\pi }{6}} 16 {\sin }^2\theta {\cos }^2\theta d\theta = 4 {\int }_{-\frac{\pi }{6}}^{\frac{\pi }{6}} {\sin }^2 2\theta d\theta$

$=2{\int }_{-\frac{\pi }{6}}^{\frac{\pi }{6}} (1 -\cos 4\theta ) d\theta =\int \left[\theta -\frac{\sin \theta }{4}\right]{ }_{-\frac{\pi }{6}}^{\frac{\pi }{6}}$

$=2\left[\frac{\pi }{3}-\frac{\pi }{4}\left\{\frac{\sqrt{3}}{2}-\left(-\frac{\sqrt{3}}{2}\right)\right\}\right]=\frac{2\pi }{3}-\frac{\sqrt{3}}{2}$

$\sqrt{{\left(x-0\right)}^2+{\left(y-2\right)}^2}=\frac{1}{2}\left|\frac{y+4}{1}\right|$

$\Rightarrow x^2+y^2+4-4y=\frac{1}{4}\left(y^2+16+8y\right)$

$\Rightarrow 4x^2+4y^2-16y+16= y^2+8y+16$

$\Rightarrow 4x^2+3y^2-24y=0$

$4x^2+3(y^2-8y+4^2) =48$

$\Rightarrow \frac{x^2}{12}+\frac{{\left(y-4\right)}^2}{4^2}=1$ এখানে, $4^2>12$

অতএব, উপকেন্দ্রিক দৈর্ঘ্য $=\frac{2\times 12}{4}=6$ একক 

$\underline{a}\times \underline{b}=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 2& 1& -3\\ 1& -2& 1\end{array}\right|$

$=\overbrace{i}(1-6)-\overbrace{j}(2+3)+\overbrace{k}(-4-1) = -5\overbrace{i}-5\overbrace{j}-5\overbrace{k}$

অতএব, $\overbrace{\eta }=\frac{\underline{a}\times \underline{b}}{\left|\underline{a} \times \right|\underline{b}} =\pm \frac{-5\hat{i}-5\hat{j}-5\hat{k}}{\sqrt[5]{3}}$অতএব, $\overbrace{\eta } =\pm \frac{1}{\sqrt{3}}\left(\overbrace{i}+\overbrace{j}+\overbrace{k}\right)$

অতএব, নির্ণেয় ভেক্টর $=5\overbrace{\eta }=\pm \frac{5}{\sqrt{3}}\left(\overbrace{i}+\overbrace{j}+\overbrace{k}\right)$

 $x^2+y^2+2gx+2fy+c=0$ বৃত্তটি (-1,-1) এবং (3,2) বিন্দু দিয়ে অতিক্রম করে পাই, 

2 - 2g - 2f + c = 0 …. …. …. (i) 

13 + 6g + 4f + c = 0 …. ….. … (ii) 

x² + y² -6x -4y -7 = 0 বৃত্তের (1, -2) বিন্দুতে স্পর্শক, 

x - 2y - 3 (x + 1) -2 (y- 2) - 7 = 0 $\Rightarrow x + 2y + 3=0 .... .... \left(iii\right)$

(iii) এর উপরে নির্ণেয় বৃত্তের কেন্দ্র (-g, -f0) অবস্থিত। 

$-g-2f+3=0\Rightarrow g+2f=3 .... .... ...\left(iv\right)$

(i), (ii), (iv) সমাধান করে, g = -4, $f=\frac{7}{2}, c=-3$

অতএব, বৃত্তের সমীকরণ, x² + y² - 8x + 7y - 3 = 0 

$\cos x+\cos y=a\Rightarrow 2 \cos \frac{x+y}{2}\cos \frac{x-Y}{2}=a ... ..... \left(i\right)$

$\sin x+\sin y=b\Rightarrow 2 \sin \frac{x+y}{2}\cos \frac{x-y}{2}=b .... .... \left(i\right)$

${\left[\left(i\right)\right]}^2÷{\left[\left(ii\right)\right]}^2$ করে পাই $\frac{{\cos }^2{\displaystyle \frac{x+y}{2}}}{{\sin }^2{\displaystyle \frac{x+y}{2}}}=\frac{a^2}{b^2}$

$\Rightarrow \frac{{\cos }^2{\displaystyle \frac{x+y}{2}}-{\sin }^2{\displaystyle \frac{x+y}{2}}}{{\cos }^2{\displaystyle \frac{x+y}{2}}+{\sin }^2{\displaystyle \frac{x+y}{2}}}=\frac{a^2}{b^2}$

অতএব, $\cos (x+y)=\frac{a^2-b^2}{a^2+b^2}$