সৃজনশীলঃ ‘sec⁡θ−tan⁡θ=P এবং 2cos2⁡θ+22sin⁡θ=Q.…’ সিলেট বোর্ড ২০২৪ – ত্রিকোণমিতি, উচ্চতর গণিত (এসএসসি)

দেওয়া আছে, $\sec \theta -\tan \theta =P$

এখন, $P=x$ হলে,

$\sec \theta -\tan \theta =x$

বা, $\frac{1}{\cos \theta }-\frac{\sin \theta }{\cos \theta }=x[\because \sec \theta =\frac{1}{\cos \theta }$ এবং $\tan \theta =\frac{\sin \theta }{\cos \theta }]$

বা, $\frac{1-\sin \theta }{\cos \theta }=x$

বা, $\frac{(1-\sin \theta {)}^2}{{\cos }^2\theta }=x^2$ [বর্গ করে]

বা, $\frac{(1-\sin \theta {)}^2}{1-{\sin }^2\theta }=x^2[\because {\cos }^2\theta =1-{\sin }^2\theta ]$

বা, $\frac{(1-\sin \theta )(1-\sin \theta )}{(1+\sin \theta )(1-\sin \theta )}=x^2$

বা, $\frac{1-\sin \theta }{1+\sin \theta }=x^2$

বা, $\frac{1-\sin \theta +1+\sin \theta }{1-\sin \theta -1-\sin \theta }=\frac{x^2+1}{x^2-1}$ [যোজন-বিয়োজন করে]

বা, $\frac{2}{-2\sin \theta }=\frac{1+x^2}{-(1-x^2)}$ বা, $\frac{1}{\sin \theta }=\frac{1+x^2}{1-x^2}$

বা, $\mathrm{cosec}\theta =\frac{1+x^2}{1-x^2}[\because \frac{1}{\sin \theta }=\mathrm{cosec}\theta ]$

$\therefore \mathrm{cosec}\theta =\frac{1+x^2}{1-x^2}$ (প্রমাণিত)

দেওয়া আছে, $2{\cos }^2\theta +2\sqrt{2}\sin \theta =Q$

এখন, $Q=3$ হলে, $2{\cos }^2\theta +2\sqrt{2}\sin \theta =3$

বা, $2(1-{\sin }^2\theta )+2\sqrt{2}\sin \theta -3=0$

বা, $2-2{\sin }^2\theta +2\sqrt{2}\sin \theta -3=0$

বা, $-2{\sin }^2\theta +2\sqrt{2}\sin \theta -1=0$

বা, $-(2{\sin }^2\theta -2\sqrt{2}\sin \theta +1)=0$

বা, $2{\sin }^2\theta -2\sqrt{2}\sin \theta +1=0$

বা, $(\sqrt{2}\sin \theta {)}^2-2\sqrt{2}\sin \theta \cdot 1+1^2=0$

বা, $(\sqrt{2}\sin \theta -1{)}^2=0$

বা, $\sqrt{2}\sin \theta -1=0$

বা, $\sqrt{2}\sin \theta =1$

বা, $\sin \theta =\frac{1}{\sqrt{2}}$

প্রথম চতুর্ভাগে, $\sin \theta =\frac{1}{\sqrt{2}}=\sin \frac{\pi }{4}\therefore \theta =\frac{\pi }{4}$

দ্বিতীয় চতুর্ভাগে, $\sin \theta =\frac{1}{\sqrt{2}}=\sin (\pi -\frac{\pi }{4})=\sin \frac{3\pi }{4}\therefore \theta =\frac{3\pi }{4}$

নির্ণেয় মান: $\theta =\frac{\pi }{4},\frac{3\pi }{4}$.