দেওয়া আছে,

$ A=\sin ^{-1} \frac{2}{3}, B=\cos ^{-1} \frac{3}{4}, C=\tan ^{-1} \frac{1}{\sqrt{5}} $

ধরি, $B=\cos ^{-1} \frac{3}{4}=\theta$

বা, $\cos \theta=\frac{3}{4}$

$ \therefore \sin \theta=\frac{\sqrt{7}}{4} $

$\tan \frac{\theta}{2}=\frac{\sin \frac{\theta}{2}}{\cos \frac{\theta}{2}}=\frac{2 \sin \frac{\theta}{2} \cos \frac{\theta}{2}}{2 \cos ^2 \frac{\theta}{2}}=\frac{\sin \theta}{1+\cos \theta}=\frac{\frac{\sqrt{7}}{4}}{1+\frac{3}{4}}=\frac{\frac{\sqrt{7}}{4}}{\frac{7}{4}}$

$\therefore \tan \frac{\theta}{2}=\frac{1}{\sqrt{7}}$

বা, $\frac{\theta}{2}=\tan ^{-1} \frac{1}{\sqrt{7}}$

$ \therefore \frac{1}{2} \cos ^{-1} \frac{3}{4}=\tan ^{-1} \frac{1}{\sqrt{7}} $

বামপক্ষ $ =A-\frac{1}{2} B+C $

$ =\sin ^{-1} \frac{2}{3}-\frac{1}{2} \cos ^{-1} \frac{3}{4}+\tan ^{-1} \frac{1}{\sqrt{5}} $

$ =\sin ^{-1} \frac{2}{3}-\tan ^{-1} \frac{1}{\sqrt{7}}+\tan ^{-1} \frac{1}{\sqrt{5}} $

$ =\tan ^{-1} \frac{2}{\sqrt{5}}+\tan ^{-1} \frac{1}{\sqrt{5}}-\tan ^{-1} \frac{1}{\sqrt{7}}$

$ =\tan ^{-1} \frac{\frac{2}{\sqrt{5}}+\frac{1}{\sqrt{5}}}{1-\frac{2}{\sqrt{5}} \cdot \frac{1}{\sqrt{5}}}-\tan ^{-1} \frac{1}{\sqrt{7}} $

$ =\tan ^{-1} \frac{\frac{3}{\sqrt{5}}}{\frac{5-2}{5}}-\tan ^{-1} \frac{1}{\sqrt{7}} $

$ =\tan ^{-1} \sqrt{5}-\tan ^{-1} \frac{1}{\sqrt{7}}=\tan ^{-1} \frac{\sqrt{5}-\frac{1}{\sqrt{7}}}{1+\sqrt{5} \cdot \frac{1}{\sqrt{7}}} $

$ =\tan ^{-1} \frac{\frac{\sqrt{35}-1}{\sqrt{7}}}{\frac{\sqrt{7}+\sqrt{5}}{\sqrt{7}}}=\tan ^{-1}\left(\frac{\sqrt{35}-1}{\sqrt{7}} \times \frac{\sqrt{7}}{\sqrt{7}+\sqrt{5}}\right) $

$ =\tan ^{-1}\left(\frac{\sqrt{\frac{7}{35}}-1}{\sqrt{7}+\sqrt{5}}\right)= $ ডানপক্ষ

$ \therefore A-\frac{1}{2} B+C=\tan ^{-1}\left(\frac{\sqrt{35}-1}{\sqrt{7}+\sqrt{5}}\right) $ (প্রমাণিত)

Chapter 7: বিপরীত ত্রিকোনমিতিক ফাংশন ও ত্রিকোণমিতিক সমীকরণ — CQs