সৃজনশীলঃ ‘দৃশ্যকল্প-১ : f(x)=sin⁡2xদৃশ্যকল্প-২…’ আনন্দ মোহন কলেজ ২০২৬ – ফাংশন ও ফাংশনের লেখচিত্র, উচ্চতর গণিত, উচ্চতর গণিত ১ম পত্র (এইচএসসি)

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দেওয়া আছে, $\sqrt{2}\cos A=\cos B+{\cos }^{3}B$ ......... (i)

এবং $\sqrt{2}\sin A=\sin B-{\sin }^{3}B$ ......... (ii)

(i) ও (ii) কে বর্গ করে যোপ করে, পাই,

$2({\cos }^{2}A÷{\sin }^{2}A)={\cos }^{2}B÷{\sin }^{2}B÷2({\cos }^{4}B-{\sin }^{4}B)÷{\cos }^{6}B+{\sin }^{6}B$

বা, $2\cdot 1=1+2\{{({\cos }^{2}B)}^2-{({\sin }^{2}B)}^2\}+{({\cos }^{2}B)}^3+{({\sin }^{2}B)}^3$

বা, $2=1+2({\cos }^{2}B+{\sin }^{2}B)({\cos }^{2}B-{\sin }^{2}B)$

$+{({\cos }^{2}B+{\sin }^{2}B)}^3-3{\cos }^{2}B\cdot {\sin }^{2}B({\cos }^{2}B+{\sin }^{2}B)$

বা, $1=2.1({\cos }^{2}B-{\sin }^{2}B)+1^3÷3{\cos }^{2}B\cdot {\sin }^{2}B$

বা, $3{\cos }^{2}B(1-{\cos }^{2}B)-2({\cos }^{2}B-1+{\cos }^{2}B)=0$

বা, $3{\cos }^{2}B-3{\cos }^{4}B-2{\cos }^{2}B+2-2{\cos }^{2}B=0$

বা, $3{\cos }^{4}B+{\cos }^{2}B-2=0$

বা, $3{\cos }^{4}B-2{\cos }^{2}B+3{\cos }^{2}B-2=0$

বা, ${\cos }^{2}B(3{\cos }^{2}B-2)+1(3{\cos }^{2}B-2)=0$

বা, $(3{\cos }^{2}B-2)({\cos }^{2}B+1)=0$

বা, $3{\cos }^{2}B-2=0[\because {\cos }^{2}B+1\ne 0]$

বা, $3{\cos }^{2}B=2$

বা, ${\cos }^{2}B=\frac{2}{3}$

$\therefore \cos B=\pm \sqrt{\frac{2}{3}}$

এখন, $\sin (A-B)=\sin A\cos B-\cos A\sin B$

$=\cos B\cdot \frac{1}{\sqrt{2}}(\sin B-{\sin }^{3}B)-\sin B\frac{1}{\sqrt{2}}(\cos B+{\cos }^{3}B)$ [(i) ও (ii) থেকে]

$=-\frac{1}{\sqrt{2}}\cos B{\sin }^{3}B-\frac{1}{\sqrt{2}}\sin B{\cos }^{3}B$

$=-\frac{1}{\sqrt{2}}\sin B\cos B({\sin }^{2}B+{\cos }^{2}B)$

$=-\frac{1}{\sqrt{2}}\sin B\cos B\times 1=-\frac{1}{\sqrt{2}}\cos B\sqrt{1-{\cos }^{2}B}$

$=-\frac{1}{\sqrt{2}}(\pm \sqrt{\frac{2}{3}})\sqrt{1-{(\pm \sqrt{\frac{2}{3}})}^2}[\cos B-$ এর মান বসিয়ে]

$=\pm \frac{1}{\sqrt{2}}\cdot \frac{\sqrt{2}}{\sqrt{3}}\cdot \frac{1}{\sqrt{3}}=\pm \frac{1}{3}$

বা, $\sin (A-B)=\pm \frac{1}{3}$

বা, $\frac{1}{\mathrm{cosec}(A-B)}=\pm \frac{1}{3}$

$\therefore \mathrm{cosec}(A-B)=\pm 3$. (প্রমানিত)