বিপরীত ত্রিকোণমিতিক ফাংশন ও পর্যায়

tan12sinα2sin(π4β2)cosα2cos(π4β2)cos2α2cos2(π4β2)sin2α2sin2(π4β2)=? \tan ^{-1} \frac{2 \sin \frac{\alpha}{2} \sin \left(\frac{\pi}{4}-\frac{\beta}{2}\right) \cos \frac{\alpha}{2} \cos \left(\frac{\pi}{4}-\frac{\beta}{2}\right)}{\cos ^{2} \frac{\alpha}{2} \cos ^{2}\left(\frac{\pi}{4}-\frac{\beta}{2}\right)-\sin ^{2} \frac{\alpha}{2} \sin ^{2}\left(\frac{\pi}{4}-\frac{\beta}{2}\right)}= ?

Solve:=tan12sinα2sin(π4β2)cosα2cos(π4β2)cos2α2cos2(π4β2)sin2α2sin2(π4β2) now, 2sinα2sin(π4β2)cosα2cos(π4β2)=12sinαsin2(π4β2)=12sinαsin(π2β)=12sinαcosβ , and cos2α2cos2(π4β2)sin2α2sin2(π4β2)=cosα2cos(π4β2)+sinα2sin(π4β2)}{cosα2cos(π4β2)sinα2sin(π4β2)}=cos(α2+β2π4)cos(α2β2+π4)=cos{α2+(β2π4)}cos{α2(β2π4)}=12{cos2α2+cos2(β2π4)}=12{cosα+cos(βπ2)}=12{cosα+sinβ)=tan12sinα2sin(π4β2)cosα2cos(π4β2)cos2α2cos2(π4β2)sin2α2sin2(π4β2)=tan1sinαcosβcosα+sinβ\begin{array}{l}=\tan ^{-1} \frac{2 \sin \frac{\alpha}{2} \sin \left(\frac{\pi}{4}-\frac{\beta}{2}\right) \cos \frac{\alpha}{2} \cos \left(\frac{\pi}{4}-\frac{\beta}{2}\right)}{\cos ^{2} \frac{\alpha}{2} \cos ^{2}\left(\frac{\pi}{4}-\frac{\beta}{2}\right)-\sin ^{2} \frac{\alpha}{2} \sin ^{2}\left(\frac{\pi}{4}-\frac{\beta}{2}\right)} \\\text { now, } 2 \sin \frac{\alpha}{2} \sin \left(\frac{\pi}{4}-\frac{\beta}{2}\right) \cos \frac{\alpha}{2} \cos \left(\frac{\pi}{4}-\frac{\beta}{2}\right) \\=\frac{1}{2} \sin \alpha \sin 2\left(\frac{\pi}{4}-\frac{\beta}{2}\right) \\=\frac{1}{2} \sin \alpha \sin \left(\frac{\pi}{2}-\beta\right) \\=\frac{1}{2} \sin \alpha \cos \beta\text { , and } \\\cos ^{2} \frac{\alpha}{2} \cos ^{2}\left(\frac{\pi}{4}-\frac{\beta}{2}\right)-\sin ^{2} \frac{\alpha}{2} \sin ^{2}\left(\frac{\pi}{4}-\frac{\beta}{2}\right) \\ \left.=\cos \frac{\alpha}{2} \cos \left(\frac{\pi}{4}-\frac{\beta}{2}\right)+\sin \frac{\alpha}{2} \sin \left(\frac{\pi}{4}-\frac{\beta}{2}\right)\right\} \\ \left\{\cos \frac{\alpha}{2} \cos \left(\frac{\pi}{4}-\frac{\beta}{2}\right)-\sin \frac{\alpha}{2} \sin \left(\frac{\pi}{4}-\frac{\beta}{2}\right)\right\} \\ =\cos \left(\frac{\alpha}{2}+\frac{\beta}{2}-\frac{\pi}{4}\right) \cdot \cos \left(\frac{\alpha}{2}-\frac{\beta}{2}+\frac{\pi}{4}\right) \\ =\cos \left\{\frac{\alpha}{2}+\left(\frac{\beta}{2}-\frac{\pi}{4}\right)\right\} \cdot \cos \left\{\frac{\alpha}{2}-\left(\frac{\beta}{2}-\frac{\pi}{4}\right)\right\} \\ =\frac{1}{2}\left\{\cos 2 \cdot \frac{\alpha}{2}+\cos 2 \cdot\left(\frac{\beta}{2}-\frac{\pi}{4}\right)\right\} \\ =\frac{1}{2}\left\{\cos \alpha+\cos \left(\beta-\frac{\pi}{2}\right)\right\} \\ =\frac{1}{2}\{\cos \alpha+\sin \beta) \\=\tan ^{-1} \frac{2 \sin \frac{\alpha}{2} \sin \left(\frac{\pi}{4}-\frac{\beta}{2}\right) \cos \frac{\alpha}{2} \cos \left(\frac{\pi}{4}-\frac{\beta}{2}\right)}{\cos ^{2} \frac{\alpha}{2} \cos ^{2}\left(\frac{\pi}{4}-\frac{\beta}{2}\right)-\sin ^{2} \frac{\alpha}{2} \sin ^{2}\left(\frac{\pi}{4}-\frac{\beta}{2}\right)} = tan^{-1} \frac{sin \alpha cos \beta}{cos \alpha + sin \beta}\end{array}

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