সৃজনশীলঃ ‘Sinθ=35 এবং A=π12…’ ঝালকাঠি সরকারি কলেজ ২০২৫ – সংযুক্ত কোনের ত্রিকোনমিতিক অনুপাত, উচ্চতর গণিত, উচ্চতর গণিত ১ম পত্র (এইচএসসি)

দেওয়া আছে,

$\sin \theta =\frac{3}{5}$

$\therefore \mathrm{cosec}\theta =\frac{5}{3}$

${\sin }^2\theta =\frac{9}{25}$

$\Rightarrow 1-{\cos }^2\theta =\frac{9}{25}$

$\Rightarrow {\cos }^2\theta =1-\frac{9}{25}$

$\Rightarrow {\cos }^2\theta =\frac{16}{25}$

$\therefore \cos \theta =\pm \frac{4}{5}$

$\because \frac{\pi }{2}<\theta <H$

$\therefore \cos \theta =-\frac{4}{5}$

$\therefore \tan \theta =\frac{\sin \theta }{\cos \theta }=\frac{\frac{3}{5}}{\frac{-4}{5}}=-\frac{3}{4}$

এবং $cot\theta =-\frac{4}{3}$

এখন, $\frac{\cot \theta +\cos (-\theta )}{\mathrm{cosec}(-\theta )+\tan \theta }=\frac{\cot \theta +\cos \theta }{-\mathrm{cosec}\theta +\tan \theta }$

$=\frac{-\frac{4}{3}-\frac{4}{5}}{-\frac{5}{3}-\frac{3}{4}}$

$=\frac{-{(\frac{4}{3}+\frac{4}{5})}^4}{-(-5+3)}$

$\frac{20+12}{15}$

$=\frac{20+9}{12}$

$=\frac{32}{15}\times \frac{12}{29}$

$=\frac{128}{145}$ (Ans) 

L.H.S $=\tan A\tan 3A\tan 5A\tan 7A\tan 11A$

যেখান, $A=\frac{\pi }{12}$

$A={15}^{\circ }$

$\therefore \text{ L.H.S }=\tan {15}^{\circ }\tan {45}^{\circ }\tan {75}^{\circ }\tan {105}^{\circ }\tan {165}^{\circ }$

$=\tan {15}^{\circ }\cdot 1\cdot \tan ({90}^{\circ }-{15}^{\circ })\tan (90+15)\tan (180-15)$

$=\tan {15}^{\circ }\cdot \cot {15}^{\circ }(-\cot {15}^{\circ })(-\tan {15}^{\circ })$

= 1

= R.H.S