সৃজনশীলঃ ‘A=sec⁡θ+tan⁡θ এবং B=sec⁡θ−tan⁡θ…’ কুমিল্লা বোর্ড ২০২৩ – ত্রিকোণমিতি, উচ্চতর গণিত (এসএসসি)

দেওয়া আছে, $A=\sec \theta +\tan \theta$

$B=\sec \theta -\tan \theta$

বামপক্ষ $=\frac{A-1}{1-B}=\frac{\cos \theta }{1-\sin \theta }=\frac{\sec \theta +\tan \theta -1}{1-\sec \theta +\tan \theta }$

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

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

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

$=\sec \theta +\tan \theta$

$=\frac{1}{\cos \theta }+\frac{\sin \theta }{\cos \theta }=\frac{1+\sin \theta }{\cos \theta }$

$=\frac{(1+\sin \theta )(1-\sin \theta )}{(1-\sin \theta )\cos \theta }$

$=\frac{1-{\sin }^2\theta }{(1-\sin \theta )\cos \theta }$

$=\frac{{\cos }^2\theta }{(1-\sin \theta )\cos \theta }$

$=\frac{\cos \theta }{1-\sin \theta }$

=  ডানপক্ষ

∴ বামপক্ষ $=$ ডানপক্ষ (প্রমাণিত)

দেওয়া আছে, $B=\sec \theta -\tan \theta =\frac{1}{\sqrt{3}}$

বা, $\frac{1}{\cos \theta }-\frac{\sin \theta }{\cos \theta }=\frac{1}{\sqrt{3}}$

বা, $\frac{1-\sin \theta }{\cos \theta }=\frac{1}{\sqrt{3}}$

বা, $\sqrt{3}(1-\sin \theta )=\cos \theta$

বা, $3(1-\sin \theta {)}^2={\cos }^2\theta$ [বর্গ করে]

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

বা, $3-6\sin \theta +3{\sin }^2\theta =1-{\sin }^2\theta$

বা, $4{\sin }^2\theta -6\sin \theta +2=0$

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

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

বা, $2\sin \theta (\sin \theta -1)-1(\sin \theta -1)=0$

বা, $(\sin \theta -1)(2\sin \theta -1)=0$

হয়, $\sin \theta -1=0$

বা, $\sin \theta =1$

কিন্তু $\sin \theta \ne 1[\therefore 0<\theta <\frac{\pi }{2}]$

অথবা, $2\sin \theta -1=0$

বা, $2\sin \theta =1$

বা, $\sin \theta =\frac{1}{2}$

বা, $\sin \theta =\sin \frac{\pi }{6}$

$\therefore \theta =\frac{\pi }{6}$

∴ নির্ণেয় $\theta =\frac{\pi }{6}$ (Ans.)