সৃজনশীলঃ ‘দৃশ্যকল্প-১: P=cos2⁡5α+cos2⁡5β−sin2⁡5γ+2cos⁡5αcos⁡5βcos⁡5γদৃশ্যকল্প-২ :…’ মাইলস্টোন কলেজ ২০২৫ – সংযুক্ত কোনের ত্রিকোনমিতিক অনুপাত, উচ্চতর গণিত, উচ্চতর গণিত ১ম পত্র (এইচএসসি)

[বি: দ্র: M এর পরিবর্তে P হবে]

দেওয়া আছে,

$M={\cos }^{2}5\alpha +{\cos }^{2}5\beta -{\sin }^{2}5\gamma +2\cos 5\alpha \cos 5\beta \cos 5\gamma$

যেখানে, $\alpha +\beta +\gamma =\pi$

এখন,

$\begin{array}{cc}& {\cos }^{2}5\alpha +{\cos }^{2}5\beta -{\sin }^{2}5\gamma =\frac{1}{2}(2{\cos }^{2}5\alpha +{\cos }^{2}5\beta )-{\sin }^{2}5\gamma \\ & =\frac{1}{2}(1+\cos 10\alpha +1+\cos 10\beta )-{\sin }^{2}5\gamma \\ & =\frac{1}{2}(2+\cos 10\alpha +\cos 10\beta )-{\sin }^{2}5\gamma \\ & =\frac{1}{2}[2+2\cos (5\alpha +5\beta )\cos (5\alpha -5\beta \left)\right]-{\sin }^{2}5\gamma \\ & =1+\cos (5\alpha +5\beta )\cos (5\alpha -5\beta )-{\sin }^{2}5\gamma \\ & =\cos 5(\alpha +\beta )\cos (5\alpha -5\beta )+{\cos }^{2}5\gamma \\ & =\cos 5(\pi -\gamma )\cos (5\alpha -5\beta )+{\cos }^{2}5\gamma [\because \alpha +\beta +\gamma =\pi ]\\ & =\cos (5\pi -5\gamma )\cos (5\alpha -5\beta )+{\cos }^{2}5\gamma \\ & =-\cos 5\gamma \cos (5\alpha -5\beta )+{\cos }^{2}5\gamma \\ & =-\cos 5\gamma [\cos (5\alpha -5\beta )-\cos 5\gamma ]\\ & =-\cos 5\gamma [\cos (5\alpha -5\beta )-\cos \{5\pi -(5\alpha +5\beta )\left\}\right]\\ & =-\cos 5\gamma [\cos (5\alpha -5\beta )+\cos (5\alpha +5\beta \left)\right]\\ & =-2\cos 5\alpha \cos 5\beta \cos 5\gamma \end{array}$

বা, ${\cos }^{2}5\alpha +{\cos }^{2}5\beta -{\sin }^{2}5\gamma =-2\cos 5\alpha \cos 5\beta \cos 5\gamma$

বা, ${\cos }^{2}5\alpha +{\cos }^{2}5\beta -{\sin }^{2}5\gamma +2\cos 5\alpha \cos 5\beta \cos 5\gamma =0$

$\therefore M=0$. (প্রমানিত)

[বি: দ্র: $\phi$ এর পরিবর্তে $\alpha$ হবে]

দেওয়া আছে,

$m=\frac{P+Q}{2}$

$\Rightarrow P+Q=2m...........\left(i\right)$

$n=\frac{P-Q}{2}$

$P-Q=2n............\left(ii\right)$

$\left(i\right)+\left(ii\right)$ হতে পাই,

$P+\theta +P-Q=2m+2n$

$\Rightarrow 2p=2(m+n)$

$\Rightarrow p=m+n$

$\left(i\right)–\left(ii\right)$ হতে পাই,

$P+Q-P+Q=2m-2n$

$\Rightarrow 2Q=2(m-n)$

$\Rightarrow Q=m-n$

দেওয়া আছে,

$\sec \theta =\frac{m-n\cos \phi }{m\cos \phi -n}$

$\Rightarrow \frac{1}{\cos \theta }=\frac{m-n\cos \phi }{m\cos \phi -n}$

$\Rightarrow m\cos \theta -n\cos \theta \cdot \cos \phi =m\cos \phi -n$

$\Rightarrow m\cos \theta -m\cos \phi =n\cos \theta \cdot \cos \phi -n$

$\Rightarrow m(\cos \theta -\cos \phi )=n(\cos \theta \cdot \cos \phi -1)$

$\Rightarrow \frac{m}{n}=\frac{\cos \theta \cdot \cos \phi -1}{\cos \theta -\cos \phi }$

$\Rightarrow \frac{m+n}{m-n}=\frac{\cos \theta \cdot \cos \phi -1+\cos \theta -\cos \phi }{\cos \theta \cdot \cos \phi -1-\cos \theta +\cos \phi }$ (যোজন - বিয়োজন করে)

$\Rightarrow \frac{m+n}{m-n}=\frac{\cos \theta (1+\cos \phi )-1(1+\cos \phi )}{\cos \phi (1+\cos \theta )-1(1+\cos \theta )}$

$\Rightarrow \frac{m+n}{m-n}=\frac{(1+\cos \phi )(\cos \theta -1)}{(1+\cos \theta )(\cos \phi -1)}$

$\Rightarrow \frac{m+n}{m-n}=\frac{-(1+\cos \phi )(1-\cos \theta )}{-(1+\cos \theta )(1-\cos \phi )}$

$\Rightarrow \frac{P}{Q}=\frac{2{\cos }^2\frac{\phi }{2}\cdot {\sin }^2\frac{\theta }{2}}{2{\cos }^2\frac{\theta }{2}\cdot {\sin }^2\frac{\phi }{2}}$

$\Rightarrow \frac{P}{Q}={\cot }^2\frac{\phi }{2}\cdot {\tan }^2\frac{\theta }{2}$

$\Rightarrow \frac{P}{Q}=\frac{{\tan }^2\frac{\theta }{2}}{{\tan }^2\frac{\phi }{2}}$

$\Rightarrow \frac{\sqrt{P}}{\sqrt{Q}}=\frac{{\tan }^2\frac{\theta }{2}}{\tan \frac{\phi }{2}}$

$\Rightarrow \tan \frac{\theta }{2}\cdot \sqrt{Q}=\tan \frac{\phi }{2}\sqrt{P}$

$\Rightarrow \frac{\tan \frac{\theta }{2}}{\sqrt{P}}=\frac{\tan \frac{\phi }{2}}{\sqrt{Q}}$ (Proved)