বিপরীত ত্রিকোণমিতিক ফাংশন ও পর্যায়
tan−12sinα2sin(π4−β2)cosα2cos(π4−β2)cos2α2cos2(π4−β2)−sin2α2sin2(π4−β2)=? \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−1cos22αcos2(4π−2β)−sin22αsin2(4π−2β)2sin2αsin(4π−2β)cos2αcos(4π−2β)=?
tan−1sinαcosβ\tan^{-1} \frac{\sin \alpha}{\cos \beta}tan−1cosβsinα
tan−1cosαsinβ\tan^{-1} \frac{\cos \alpha}{\sin \beta}tan−1sinβcosα
tan−1sinβcosα\tan^{-1} \frac{\sin \beta}{\cos \alpha}tan−1cosαsinβ
tan−1sinαcosβcosα+sinβ\tan^{-1} \frac{\sin \alpha \cos \beta}{\cos \alpha + \sin \beta}tan−1cosα+sinβsinαcosβ
Solve:=tan−12sinα2sin(π4−β2)cosα2cos(π4−β2)cos2α2cos2(π4−β2)−sin2α2sin2(π4−β2) now, 2sinα2sin(π4−β2)cosα2cos(π4−β2)=12sinαsin2(π4−β2)=12sinαsin(π2−β)=12sinαcosβ , and cos2α2cos2(π4−β2)−sin2α2sin2(π4−β2)=cosα2cos(π4−β2)+sinα2sin(π4−β2)}{cosα2cos(π4−β2)−sinα2sin(π4−β2)}=cos(α2+β2−π4)⋅cos(α2−β2+π4)=cos{α2+(β2−π4)}⋅cos{α2−(β2−π4)}=12{cos2⋅α2+cos2⋅(β2−π4)}=12{cosα+cos(β−π2)}=12{cosα+sinβ)=tan−12sinα2sin(π4−β2)cosα2cos(π4−β2)cos2α2cos2(π4−β2)−sin2α2sin2(π4−β2)=tan−1sinαcosβcosα+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}=tan−1cos22αcos2(4π−2β)−sin22αsin2(4π−2β)2sin2αsin(4π−2β)cos2αcos(4π−2β) now, 2sin2αsin(4π−2β)cos2αcos(4π−2β)=21sinαsin2(4π−2β)=21sinαsin(2π−β)=21sinαcosβ , and cos22αcos2(4π−2β)−sin22αsin2(4π−2β)=cos2αcos(4π−2β)+sin2αsin(4π−2β)}{cos2αcos(4π−2β)−sin2αsin(4π−2β)}=cos(2α+2β−4π)⋅cos(2α−2β+4π)=cos{2α+(2β−4π)}⋅cos{2α−(2β−4π)}=21{cos2⋅2α+cos2⋅(2β−4π)}=21{cosα+cos(β−2π)}=21{cosα+sinβ)=tan−1cos22αcos2(4π−2β)−sin22αsin2(4π−2β)2sin2αsin(4π−2β)cos2αcos(4π−2β)=tan−1cosα+sinβsinαcosβ
cotk=12 \cot{k} = \frac{1}{2} cotk=21 হলে cottan−1secsin−1cotk \cot{\tan^{- 1}{\sec{\sin^{- 1}{\cot{k}}}}} cottan−1secsin−1cotk এর মান কত?
sincot−1tancos−134 \sin \cot ^{-1} \tan \cos ^{-1} \frac{3}{4} sincot−1tancos−143 এর মান কত?
tan(sin−10.5)tan( sin^{-1} 0.5)tan(sin−10.5) এর মান কত?
i. sin−1x+cos−1x=π2 \sin ^{-1} x+\cos ^{-1} x=\frac{\pi}{2} sin−1x+cos−1x=2π
ii. tan−1x+cot−1x=π \tan ^{-1} x+\cot ^{-1} x=\pi tan−1x+cot−1x=π
iii. sec−1x+cos−1x=π2 \sec ^{-1} x+\cos ^{-1} x=\frac{\pi}{2} sec−1x+cos−1x=2π
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