যদি sin17\(^o\)=\(\cfrac{x}{y}\) হয়, তাহলে দেখাও sec17\(^o\)−sin73\(^o\)=\(\cfrac{x^2}{y\sqrt{y^2−x^2}}\)
Madhyamik 2020
\(∵ \sin17°=\cfrac{x}{y}\)
বা, \(\sin^2 17°=(\cfrac{x}{y})^2\)
বা, \(1- \sin^2 17°=1-\cfrac{x^2}{y^2}\)
বা, \(\cos^2 17°=\cfrac{y^2-x^2}{y^2}\)
বা, \(\cos 17° = \sqrt{\cfrac{y^2-x^2}{y^2}}\)
বা, \(\cos 17° = \cfrac{\sqrt{y^2-x^2}}{y}\)
\(\sec 17° - \sin 73°\)
\(=\cfrac{1}{\cos 17°} -\cfrac{1}{\sin(90°-17°)}\)
\(= \cfrac{1}{\cos 17°} -\cos 17°\)
\(=\cfrac{1}{\cfrac{\sqrt{y^2-x^2}}{y}}-\cfrac{\sqrt{y^2-x^2}}{y}\)
\(=\cfrac{y}{\sqrt{y^2-x^2}}–\cfrac{\sqrt{y^2-x^2}}{y}\)
\(=\cfrac{y^2-(y^2-x^2 )}{y\sqrt{y^2-x^2}}\)
\(=\cfrac{y^2-y^2+x^2}{y\sqrt{y^2-x^2}}\)
\(=\cfrac{x^2}{y\sqrt{y^2-x^2}}\) (প্রমাণিত)