প্রমাণঃ
ধরি,
\(\sin{\theta}=x\)
\(\Rightarrow \theta=\sin^{-1}{x}\)
আবার,
\(cosec \ {\theta}=\frac{1}{\sin{\theta}}\)
\(\Rightarrow \theta=cosec^{-1}{\left(\frac{1}{\sin{\theta}}\right)}\)
\(\therefore \sin^{-1}{x}=cosec^{-1}{\left(\frac{1}{x}\right)}\) ➜Note \(\because \sin{\theta}=x\)
\(\therefore \theta=\sin^{-1}{x}\)

\(\sin^{-1}{x}=cosec^{-1}{\left(\frac{1}{x}\right)}\)
ধরি,
\(cosec \ {\theta}=x\)
\(\Rightarrow \theta=cosec^{-1}{x}\)
আবার,
\(\sin{\theta}=\frac{1}{cosec \ {\theta}}\)
\(\Rightarrow \theta=\sin^{-1}{\left(\frac{1}{cosec \ {\theta}}\right)}\)
\(\therefore cosec^{-1}{x}=\sin^{-1}{\left(\frac{1}{x}\right)}\) ➜Note \(\because cosec \ {\theta}=x\)
\(\therefore \theta=cosec^{-1}{x}\)

\(cosec^{-1}{x}=\sin^{-1}{\left(\frac{1}{x}\right)}\)
ধরি,
\(\sin{\theta}=x\)
\(\Rightarrow \theta=\sin^{-1}{x}\)
আবার,
\(\sin^2{\theta}+\cos^2{\theta}=1\)
\(\Rightarrow \cos^2{\theta}=1-\sin^2{\theta}\)
\(\Rightarrow \cos{\theta}=\sqrt{1-\sin^2{\theta}}\)
\(\Rightarrow \theta=\cos^{-1}{\left(\sqrt{1-\sin^2{\theta}}\right)}\)
\(\therefore \sin^{-1}{x}=\cos^{-1}{\left(\sqrt{1-x^2}\right)}\) ➜Note \(\because \sin{\theta}=x\)
\(\therefore \theta=\sin^{-1}{x}\)

\(\sin^{-1}{x}=\cos^{-1}{\left(\sqrt{1-x^2}\right)}\)
ধরি,
\(\sin{\theta}=x\)
\(\Rightarrow \theta=\sin^{-1}{x}\)
আবার,
\(\sec{\theta}=\frac{1}{\cos{\theta}}\)
\(\Rightarrow \theta=\sec^{-1}{\left(\frac{1}{\cos{\theta}}\right)}\)
\(\Rightarrow \theta=\sec^{-1}{\left(\frac{1}{\sqrt{1-\sin^2{\theta}}}\right)}\) ➜Note \(\because \cos{\theta}=\sqrt{1-\sin^2{\theta}}\)

\(\therefore \sin^{-1}{x}=\sec^{-1}{\left(\frac{1}{\sqrt{1-x^2}}\right)}\) ➜Note \(\because \sin{\theta}=x\)
\(\therefore \theta=\sin^{-1}{x}\)

\(\sin^{-1}{x}=\sec^{-1}{\left(\frac{1}{\sqrt{1-x^2}}\right)}\)
ধরি,
\(\sin{\theta}=x\)
\(\Rightarrow \theta=\sin^{-1}{x}\)
আবার,
\(\tan{\theta}=\frac{\sin{\theta}}{\cos{\theta}}\)
\(\Rightarrow \theta=\tan^{-1}{\left(\frac{\sin{\theta}}{\cos{\theta}}\right)}\)
\(\Rightarrow \theta=\tan^{-1}{\left(\frac{\sin{\theta}}{\sqrt{1-\sin^2{\theta}}}\right)}\) ➜Note \(\because \cos{\theta}=\sqrt{1-\sin^2{\theta}}\)

\(\therefore \sin^{-1}{x}=\tan^{-1}{\left(\frac{x}{\sqrt{1-x^2}}\right)}\) ➜Note \(\because \sin{\theta}=x\)
\(\therefore \theta=\sin^{-1}{x}\)

\(\sin^{-1}{x}=\tan^{-1}{\left(\frac{x}{\sqrt{1-x^2}}\right)}\)
ধরি,
\(\sin{\theta}=x\)
\(\Rightarrow \theta=\sin^{-1}{x}\)
আবার,
\(\cot{\theta}=\frac{\cos{\theta}}{\sin{\theta}}\)
\(\Rightarrow \theta=\cot^{-1}{\left(\frac{\cos{\theta}}{\sin{\theta}}\right)}\)
\(\Rightarrow \theta=\cot^{-1}{\left(\frac{\sqrt{1-\sin^2{\theta}}}{\sin{\theta}}\right)}\) ➜Note \(\because \cos{\theta}=\sqrt{1-\sin^2{\theta}}\)

\(\therefore \sin^{-1}{x}=\cot^{-1}{\left(\frac{\sqrt{1-x^2}}{x}\right)}\) ➜Note \(\because \sin{\theta}=x\)
\(\therefore \theta=\sin^{-1}{x}\)

\(\sin^{-1}{x}=\cot^{-1}{\left(\frac{\sqrt{1-x^2}}{x}\right)}\)

প্রমাণঃ
ধরি,
\(\cos{\theta}=x\)
\(\Rightarrow \theta=\cos^{-1}{x}\)
আবার,
\(\sec{\theta}=\frac{1}{\cos{\theta}}\)
\(\Rightarrow \theta=\sec^{-1}{\left(\frac{1}{\cos{\theta}}\right)}\)
\(\therefore \cos^{-1}{x}=\sec^{-1}{\left(\frac{1}{x}\right)}\) ➜Note \(\because \cos{\theta}=x\)
\(\therefore \theta=\cos^{-1}{x}\)

\(\cos^{-1}{x}=\sec^{-1}{\left(\frac{1}{x}\right)}\)
ধরি,
\(\sec{\theta}=x\)
\(\Rightarrow \theta=\sec^{-1}{x}\)
আবার,
\(\cos{\theta}=\frac{1}{\sec{\theta}}\)
\(\Rightarrow \theta=\cos^{-1}{\left(\frac{1}{\sec{\theta}}\right)}\)
\(\therefore \sec^{-1}{x}=\cos^{-1}{\left(\frac{1}{x}\right)}\) ➜Note \(\because \sec{\theta}=x\)
\(\therefore \theta=\sec^{-1}{x}\)

\(\sec^{-1}{x}=\cos^{-1}{\left(\frac{1}{x}\right)}\)
ধরি,
\(\cos{\theta}=x\)
\(\Rightarrow \theta=\cos^{-1}{x}\)
আবার,
\(\sin^2{\theta}+\cos^2{\theta}=1\)
\(\Rightarrow \sin^2{\theta}=1-\cos^2{\theta}\)
\(\Rightarrow \sin{\theta}=\sqrt{1-\cos^2{\theta}}\)
\(\Rightarrow \theta=\sin^{-1}{\left(\sqrt{1-\cos^2{\theta}}\right)}\)
\(\therefore \cos^{-1}{x}=\sin^{-1}{\left(\sqrt{1-x^2}\right)}\) ➜Note \(\because \cos{\theta}=x\)
\(\therefore \theta=\cos^{-1}{x}\)

\(\cos^{-1}{x}=\sin^{-1}{\left(\sqrt{1-x^2}\right)}\)
ধরি,
\(\cos{\theta}=x\)
\(\Rightarrow \theta=\cos^{-1}{x}\)
আবার,
\(cosec \ {\theta}=\frac{1}{\sin{\theta}}\)
\(\Rightarrow \theta=cosec^{-1}{\left(\frac{1}{\sin{\theta}}\right)}\)
\(\Rightarrow \theta=cosec^{-1}{\left(\frac{1}{\sqrt{1-\cos^2{\theta}}}\right)}\) ➜Note \(\because \sin{\theta}=\sqrt{1-\cos^2{\theta}}\)

\(\therefore \cos^{-1}{x}=cosec^{-1}{\left(\frac{1}{\sqrt{1-x^2}}\right)}\) ➜Note \(\because \cos{\theta}=x\)
\(\therefore \theta=\cos^{-1}{x}\)

\(\cos^{-1}{x}=cosec^{-1}{\left(\frac{1}{\sqrt{1-x^2}}\right)}\)
ধরি,
\(\cos{\theta}=x\)
\(\Rightarrow \theta=\cos^{-1}{x}\)
আবার,
\(\tan{\theta}=\frac{\sin{\theta}}{\cos{\theta}}\)
\(\Rightarrow \theta=\tan^{-1}{\left(\frac{\sin{\theta}}{\cos{\theta}}\right)}\)
\(\Rightarrow \theta=\tan^{-1}{\left(\frac{\sqrt{1-\cos^2{\theta}}}{\cos{\theta}}\right)}\) ➜Note \(\because \sin{\theta}=\sqrt{1-\cos^2{\theta}}\)

\(\therefore \cos^{-1}{x}=\tan^{-1}{\left(\frac{\sqrt{1-x^2}}{x}\right)}\) ➜Note \(\because \cos{\theta}=x\)
\(\therefore \theta=\cos^{-1}{x}\)

\(\cos^{-1}{x}=\tan^{-1}{\left(\frac{\sqrt{1-x^2}}{x}\right)}\)
ধরি,
\(\cos{\theta}=x\)
\(\Rightarrow \theta=\cos^{-1}{x}\)
আবার,
\(\cot{\theta}=\frac{\cos{\theta}}{\sin{\theta}}\)
\(\Rightarrow \theta=\cot^{-1}{\left(\frac{\cos{\theta}}{\sin{\theta}}\right)}\)
\(\Rightarrow \theta=\cot^{-1}{\left(\frac{\cos{\theta}}{\sqrt{1-\cos^2{\theta}}}\right)}\) ➜Note \(\because \sin{\theta}=\sqrt{1-\cos^2{\theta}}\)

\(\therefore \cos^{-1}{x}=\cot^{-1}{\left(\frac{x}{\sqrt{1-x^2}}\right)}\) ➜Note \(\because \cos{\theta}=x\)
\(\therefore \theta=\cos^{-1}{x}\)

\(\cos^{-1}{x}=\cot^{-1}{\left(\frac{x}{\sqrt{1-x^2}}\right)}\)

প্রমাণঃ
ধরি,
\(\tan{\theta}=x\)
\(\Rightarrow \theta=\tan^{-1}{x}\)
আবার,
\(\cot{\theta}=\frac{1}{\tan{\theta}}\)
\(\Rightarrow \theta=\cot^{-1}{\left(\frac{1}{\tan{\theta}}\right)}\)
\(\therefore \tan^{-1}{x}=\cot^{-1}{\left(\frac{1}{x}\right)}\) ➜Note \(\because \tan{\theta}=x\)
\(\therefore \theta=\tan^{-1}{x}\)

\(\tan^{-1}{x}=\cot^{-1}{\left(\frac{1}{x}\right)}\)
ধরি,
\(\cot{\theta}=x\)
\(\Rightarrow \theta=\cot^{-1}{x}\)
আবার,
\(\tan{\theta}=\frac{1}{\cot{\theta}}\)
\(\Rightarrow \theta=\tan^{-1}{\left(\frac{1}{\cot{\theta}}\right)}\)
\(\therefore \cot^{-1}{x}=\tan^{-1}{\left(\frac{1}{x}\right)}\) ➜Note \(\because \cot{\theta}=x\)
\(\therefore \theta=\cot^{-1}{x}\)

\(\cot^{-1}{x}=\tan^{-1}{\left(\frac{1}{x}\right)}\)
ধরি,
\(\tan{\theta}=x\)
\(\Rightarrow \theta=\tan^{-1}{x}\)
আবার,
\(\sin{\theta}=\frac{\sin{\theta}}{\cos{\theta}}\times\cos{\theta}\)
\(\Rightarrow \sin{\theta}=\tan{\theta}\times\frac{1}{\sec{\theta}}\) ➜Note \(\because \frac{\sin{A}}{\cos{A}}=\tan{A}\)
এবং \(\cos{A}=\frac{1}{\sec{A}}\)

\(\Rightarrow \sin{\theta}=\frac{\tan{\theta}}{\sqrt{1+\tan^2{\theta}}}\) ➜Note \(\because \sec{A}=\sqrt{1+\tan^2{A}}\)

\(\Rightarrow \theta=\sin^{-1}{\left(\frac{\tan{\theta}}{\sqrt{1+\tan^2{\theta}}}\right)}\)
\(\therefore \tan^{-1}{x}=\sin^{-1}{\left(\frac{x}{\sqrt{1+x^2}}\right)}\) ➜Note \(\because \tan{\theta}=x\)
\(\therefore \theta=\tan^{-1}{x}\)

\(\tan^{-1}{x}=\sin^{-1}{\left(\frac{x}{\sqrt{1+x^2}}\right)}\)
ধরি,
\(\tan{\theta}=x\)
\(\Rightarrow \theta=\tan^{-1}{x}\)
আবার,
\(cosec \ {\theta}=\sqrt{1+\cot^2{\theta}}\)
\(\Rightarrow cosec \ {\theta}=\sqrt{1+\frac{1}{\tan^2{\theta}}}\) ➜Note \(\because \cot{A}=\frac{1}{\tan{A}}\)

\(\Rightarrow cosec \ {\theta}=\sqrt{\frac{1+\tan^2{\theta}}{\tan^2{\theta}}}\)
\(\Rightarrow cosec \ {\theta}=\frac{\sqrt{1+\tan^2{\theta}}}{\tan{\theta}}\)
\(\Rightarrow \theta=cosec^{-1}{\left(\frac{\sqrt{1+\tan^2{\theta}}}{\tan{\theta}}\right)}\)
\(\therefore \tan^{-1}{x}=cosec^{-1}{\left(\frac{\sqrt{1+x^2}}{x}\right)}\) ➜Note \(\because \tan{\theta}=x\)
\(\therefore \theta=\tan^{-1}{x}\)

\(\tan^{-1}{x}=cosec^{-1}{\left(\frac{\sqrt{1+x^2}}{x}\right)}\)
ধরি,
\(\tan{\theta}=x\)
\(\Rightarrow \theta=\tan^{-1}{x}\)
আবার,
\(\cos{\theta}=\frac{1}{\sec{\theta}}\)
\(\Rightarrow \cos{\theta}=\frac{1}{\sqrt{1+\tan^2{\theta}}}\) ➜Note \(\because \sec{A}=\sqrt{1+\tan^2{A}}\)

\(\Rightarrow \theta=\cos^{-1}{\left(\frac{1}{\sqrt{1+\tan^2{\theta}}}\right)}\)
\(\therefore \tan^{-1}{x}=\cos^{-1}{\left(\frac{1}{\sqrt{1+x^2}}\right)}\) ➜Note \(\because \tan{\theta}=x\)
\(\therefore \theta=\tan^{-1}{x}\)

\(\tan^{-1}{x}=\cos^{-1}{\left(\frac{1}{\sqrt{1+x^2}}\right)}\)
ধরি,
\(\tan{\theta}=x\)
\(\Rightarrow \theta=\tan^{-1}{x}\)
আবার,
\(\sec{\theta}=\sqrt{1+\tan^2{\theta}}\)
\(\Rightarrow \theta=\sec^{-1}{\left(\sqrt{1+\tan^2{\theta}}\right)}\)
\(\therefore \theta=\sec^{-1}{\left(\sqrt{1+x^2}\right)}\) ➜Note \(\because \tan{\theta}=x\)
\(\therefore \theta=\tan^{-1}{x}\)

\(\tan^{-1}{x}=\sec^{-1}{\left(\sqrt{1+x^2}\right)}\)

প্রমাণঃ
ধরি,
\(\sin{\theta}=x\)
\(\Rightarrow \theta=\sin^{-1}{x}\)
আমরা জানি,
\(\sin{\theta}=\cos{\left(\frac{\pi}{2}-\theta\right)}\)
\(\Rightarrow \cos^{-1}{\sin{\theta}}=\frac{\pi}{2}-\theta\)
\(\Rightarrow \theta+\cos^{-1}{\sin{\theta}}=\frac{\pi}{2}\)
\(\therefore \sin^{-1}{x}+\cos^{-1}{x}=\frac{\pi}{2}\) ➜Note \(\because \sin{\theta}=x\)
\(\therefore \theta=\sin^{-1}{x}\)

\(\sin^{-1}{x}+\cos^{-1}{x}=\frac{\pi}{2}\)
ধরি,
\(\tan{\theta}=x\)
\(\Rightarrow \theta=\tan^{-1}{x}\)
আমরা জানি,
\(\tan{\theta}=\cot{\left(\frac{\pi}{2}-\theta\right)}\)
\(\Rightarrow \cot^{-1}{\tan{\theta}}=\frac{\pi}{2}-\theta\)
\(\Rightarrow \theta+\cot^{-1}{\tan{\theta}}=\frac{\pi}{2}\)
\(\therefore \tan^{-1}{x}+\cot^{-1}{x}=\frac{\pi}{2}\) ➜Note \(\because \tan{\theta}=x\)
\(\therefore \theta=\tan^{-1}{x}\)

\(\tan^{-1}{x}+\cot^{-1}{x}=\frac{\pi}{2}\)
ধরি,
\(\sec{\theta}=x\)
\(\Rightarrow \theta=\sec^{-1}{x}\)
আমরা জানি,
\(\sec{\theta}=cosec \ {\left(\frac{\pi}{2}-\theta\right)}\)
\(\Rightarrow cosec^{-1}{\sec{\theta}}=\frac{\pi}{2}-\theta\)
\(\Rightarrow \theta+cosec^{-1}{\sec{\theta}}=\frac{\pi}{2}\)
\(\therefore \sec^{-1}{x}+cosec^{-1}{x}=\frac{\pi}{2}\) ➜Note \(\because \sec{\theta}=x\)
\(\therefore \theta=\sec^{-1}{x}\)

\(\sec^{-1}{x}+cosec^{-1}{x}=\frac{\pi}{2}\)

প্রমাণঃ
ধরি,
\(\sin{A}=x, \ \sin{B}=y\)
\(\Rightarrow A=\sin^{-1}{x}, \ B=\sin^{-1}{y}\)
আমরা জানি,
\(\sin{(A+B)}=\sin{A}\cos{B}+\cos{A}\sin{B}\)
\(\Rightarrow \sin{(A+B)}=\sin{A}\sqrt{1-\sin^2{B}}+\sqrt{1-\sin^2{A}}\sin{B}\) ➜Note \(\because \cos{\theta}=\sqrt{1-sin^2{\theta}}\)

\(\Rightarrow A+B=\sin^{-1}{(\sin{A}\sqrt{1-\sin^2{B}}+\sin{B}\sqrt{1-\sin^2{A}})}\)
\(\therefore \sin^{-1}{x}+\sin^{-1}{y}=\sin^{-1}{(x\sqrt{1-y^2}+y\sqrt{1-x^2})}\) ➜Note \(\because \sin{A}=x, \ \sin{B}=y\)
\(\therefore A=\sin^{-1}{x}, \ B=\sin^{-1}{y}\)

\(\sin^{-1}{x}+\sin^{-1}{y}=\sin^{-1}{(x\sqrt{1-y^2}+y\sqrt{1-x^2})}\)
ধরি,
\(\sin{A}=x, \ \sin{B}=y\)
\(\Rightarrow A=\sin^{-1}{x}, \ B=\sin^{-1}{y}\)
আমরা জানি,
\(\sin{(A-B)}=\sin{A}\cos{B}-\cos{A}\sin{B}\)
\(\Rightarrow \sin{(A-B)}=\sin{A}\sqrt{1-\sin^2{B}}-\sqrt{1-\sin^2{A}}\sin{B}\) ➜Note \(\because \cos{\theta}=\sqrt{1-sin^2{\theta}}\)

\(\Rightarrow A-B=\sin^{-1}{(\sin{A}\sqrt{1-\sin^2{B}}-\sin{B}\sqrt{1-\sin^2{A}})}\)
\(\therefore \sin^{-1}{x}-\sin^{-1}{y}=\sin^{-1}{(x\sqrt{1-y^2}-y\sqrt{1-x^2})}\) ➜Note \(\because \sin{A}=x, \ \sin{B}=y\)
\(\therefore A=\sin^{-1}{x}, \ B=\sin^{-1}{y}\)

\(\sin^{-1}{x}-\sin^{-1}{y}=\sin^{-1}{(x\sqrt{1-y^2}-y\sqrt{1-x^2})}\)

প্রমাণঃ
ধরি,
\(\cos{A}=x, \ \cos{B}=y\)
\(\Rightarrow A=\cos^{-1}{x}, \ B=\cos^{-1}{y}\)
আমরা জানি,
\(\cos{(A+B)}=\cos{A}\cos{B}-\sin{A}\sin{B}\)
\(\Rightarrow \cos{(A+B)}=\cos{A}\cos{B}-\sqrt{1-\cos^2{A}}\sqrt{1-\cos^2{B}}\) ➜Note \(\because \sin{\theta}=\sqrt{1-cos^2{\theta}}\)

\(\Rightarrow A+B=\cos^{-1}{\{\cos{A}\cos{B}-\sqrt{(1-\cos^2{A})(1-\cos^2{B})}\}}\)
\(\therefore \cos^{-1}{x}+\cos^{-1}{y}=\cos^{-1}{\{xy-\sqrt{(1-x^2)(1-y^2)}\}}\) ➜Note \(\because \cos{A}=x, \ \cos{B}=y\)
\(\therefore A=\cos^{-1}{x}, \ B=\cos^{-1}{y}\)

\(\cos^{-1}{x}+\cos^{-1}{y}=\cos^{-1}{\{xy-\sqrt{(1-x^2)(1-y^2)}\}}\)
ধরি,
\(\cos{A}=x, \ \cos{B}=y\)
\(\Rightarrow A=\cos^{-1}{x}, \ B=\cos^{-1}{y}\)
আমরা জানি,
\(\cos{(A-B)}=\cos{A}\cos{B}+\sin{A}\sin{B}\)
\(\Rightarrow \cos{(A-B)}=\cos{A}\cos{B}+\sqrt{1-\cos^2{A}}\sqrt{1-\cos^2{B}}\) ➜Note \(\because \sin{\theta}=\sqrt{1-cos^2{\theta}}\)

\(\Rightarrow A-B=\cos^{-1}{\{\cos{A}\cos{B}+\sqrt{(1-\cos^2{A})(1-\cos^2{B})}\}}\)
\(\therefore \cos^{-1}{x}-\cos^{-1}{y}=\cos^{-1}{\{xy+\sqrt{(1-x^2)(1-y^2)}\}}\) ➜Note \(\because \cos{A}=x, \ \cos{B}=y\)
\(\therefore A=\cos^{-1}{x}, \ B=\cos^{-1}{y}\)

\(\cos^{-1}{x}-\cos^{-1}{y}=\cos^{-1}{\{xy+\sqrt{(1-x^2)(1-y^2)}\}}\)

প্রমাণঃ
ধরি,
\(\tan{A}=x, \ \tan{B}=y\)
\(\Rightarrow A=\tan^{-1}{x}, \ B=\tan^{-1}{y}\)
আমরা জানি,
\(\tan{(A+B)}=\frac{\tan{A}+\tan{B}}{1-\tan{A}\tan{B}}\)
\(\Rightarrow A+B=\tan^{-1}{\left(\frac{\tan{A}+\tan{B}}{1-\tan{A}\tan{B}}\right)}\)
\(\therefore \tan^{-1}{x}+\tan^{-1}{y}=\tan^{-1}{\left(\frac{x+y}{1-xy}\right)}\) ➜Note \(\because \tan{A}=x, \ \tan{B}=y\)
\(\therefore A=\tan^{-1}{x}, \ B=\tan^{-1}{y}\)

\(\tan^{-1}{x}+\tan^{-1}{y}=\tan^{-1}{\left(\frac{x+y}{1-xy}\right)}\)
ধরি,
\(\tan{A}=x, \ \tan{B}=y\)
\(\Rightarrow A=\tan^{-1}{x}, \ B=\tan^{-1}{y}\)
আমরা জানি,
\(\tan{(A-B)}=\frac{\tan{A}-\tan{B}}{1+\tan{A}\tan{B}}\)
\(\Rightarrow A-B=\tan^{-1}{\left(\frac{\tan{A}-\tan{B}}{1+\tan{A}\tan{B}}\right)}\)
\(\therefore \tan^{-1}{x}-\tan^{-1}{y}=\tan^{-1}{\left(\frac{x-y}{1+xy}\right)}\) ➜Note \(\because \tan{A}=x, \ \tan{B}=y\)
\(\therefore A=\tan^{-1}{x}, \ B=\tan^{-1}{y}\)

\(\tan^{-1}{x}-\tan^{-1}{y}=\tan^{-1}{\left(\frac{x-y}{1+xy}\right)}\)
ধরি,
\(\tan{A}=x, \ \tan{B}=y, \ \tan{C}=z\)
\(\Rightarrow A=\tan^{-1}{x}, \ B=\tan^{-1}{y}, \ C=\tan^{-1}{z}\)
আমরা জানি,
\(\tan{(A+B+C)}=\frac{\tan{A}+\tan{B}+\tan{C}-\tan{A}\tan{B}\tan{C}}{1-\tan{B}\tan{C}-\tan{C}\tan{A}-\tan{A}\tan{B}}\)
\(\Rightarrow A+B+C=\tan^{-1}{\left(\frac{\tan{A}+\tan{B}+\tan{C}-\tan{A}\tan{B}\tan{C}}{1-\tan{B}\tan{C}-\tan{C}\tan{A}-\tan{A}\tan{B}}\right)}\)
\(\therefore \tan^{-1}{x}+\tan^{-1}{y}+\tan^{-1}{z}=\tan^{-1}{\left(\frac{x+y+z-xyz}{1-yz-zx-xy}\right)}\) ➜Note \(\because \tan{A}=x, \ \tan{B}=y, \ \tan{C}=z\)
\(\therefore A=\tan^{-1}{x}, \ B=\tan^{-1}{y}, \ C=\tan^{-1}{z}\)

\(\tan^{-1}{x}+\tan^{-1}{y}+\tan^{-1}{z}=\tan^{-1}{\left(\frac{x+y+z-xyz}{1-yz-zx-xy}\right)}\)

প্রমাণঃ
ধরি,
\(\cot{A}=x, \ \cot{B}=y\)
\(\Rightarrow A=\cot^{-1}{x}, \ B=\cot^{-1}{y}\)
আমরা জানি,
\(\cot{(A+B)}=\frac{\cot{A}\cot{B}-1}{\cot{B}+\cot{A}}\)
\(\Rightarrow A+B=\cot^{-1}{\left(\frac{\cot{A}\cot{B}-1}{\cot{B}+\cot{A}}\right)}\)
\(\therefore \cot^{-1}{x}+\cot^{-1}{y}=\cot^{-1}{\left(\frac{xy-1}{y+x}\right)}\) ➜Note \(\because \cot{A}=x, \ \cot{B}=y\)
\(\therefore A=\cot^{-1}{x}, \ B=\cot^{-1}{y}\)

\(\cot^{-1}{x}+\cot^{-1}{y}=\cot^{-1}{\left(\frac{xy-1}{y+x}\right)}\)
ধরি,
\(\cot{A}=x, \ \cot{B}=y\)
\(\Rightarrow A=\cot^{-1}{x}, \ B=\cot^{-1}{y}\)
আমরা জানি,
\(\cot{(A-B)}=\frac{\cot{A}\cot{B}+1}{\cot{B}-\cot{A}}\)
\(\Rightarrow A-B=\cot^{-1}{\left(\frac{\cot{A}\cot{B}+1}{\cot{B}-\cot{A}}\right)}\)
\(\therefore \cot^{-1}{x}-\cot^{-1}{y}=\cot^{-1}{\left(\frac{xy+1}{y-x}\right)}\) ➜Note \(\because \cot{A}=x, \ \cot{B}=y\)
\(\therefore A=\cot^{-1}{x}, \ B=\cot^{-1}{y}\)

\(\cot^{-1}{x}-\cot^{-1}{y}=\cot^{-1}{\left(\frac{xy+1}{y-x}\right)}\)
ধরি,
\(\cot{A}=x, \ \cot{B}=y, \ \cot{C}=z\)
\(\Rightarrow A=\cot^{-1}{x}, \ B=\cot^{-1}{y}, \ C=\cot^{-1}{z}\)
আমরা জানি,
\(\cot{(A+B+C)}=\frac{\cot{A}\cot{B}\cot{C}-\cot{A}-\cot{B}-\cot{C}}{\cot{B}\cot{C}+\cot{C}\cot{A}+\cot{A}\cot{B}-1}\)
\(\Rightarrow A+B+C=\cot^{-1}{\left(\frac{\cot{A}\cot{B}\cot{C}-\cot{A}-\cot{B}-\cot{C}}{\cot{B}\cot{C}+\cot{C}\cot{A}+\cot{A}\cot{B}-1}\right)}\)
\(\therefore \cot^{-1}{x}+\cot^{-1}{y}+\cot^{-1}{z}=\cot^{-1}{\left(\frac{xyz-x-y-z}{yz+zx+xy-1}\right)}\) ➜Note \(\because \cot{A}=x, \ \cot{B}=y, \ \cot{C}=z\)
\(\therefore A=\cot^{-1}{x}, \ B=\cot^{-1}{y}, \ C=\cot^{-1}{z}\)

\(\cot^{-1}{x}+\cot^{-1}{y}+\cot^{-1}{z}=\cot^{-1}{\left(\frac{xyz-x-y-z}{yz+zx+xy-1}\right)}\)

প্রমাণঃ
ধরি,
\(\sin{\theta}=x\)
\(\Rightarrow \theta=\sin^{-1}{x}\)
আবার,
\(\sin{2\theta}=2\sin{\theta}\cos{\theta}\)
\(\Rightarrow \sin{2\theta}=2\sin{\theta}\sqrt{1-\sin^2{\theta}}\)
\(\Rightarrow 2\theta=\sin^{-1}{(2\sin{\theta}\sqrt{1-\sin^2{\theta}})}\)
\(\therefore 2\sin^{-1}{x}=\sin^{-1}{(2x\sqrt{1-x^2})}\) ➜Note \(\because \sin{\theta}=x\)
\(\therefore \theta=\sin^{-1}{x}\)

\(2\sin^{-1}{x}=\sin^{-1}{(2x\sqrt{1-x^2})}\)
ধরি,
\(\cos{\theta}=x\)
\(\Rightarrow \theta=\cos^{-1}{x}\)
আবার,
\(\cos{2\theta}=2\cos^2{\theta}-1\)
\(\Rightarrow 2\theta=\cos^{-1}{(2\cos^2{\theta}-1)}\)
\(\therefore 2\cos^{-1}{x}=\cos^{-1}{(2x^2-1)}\) ➜Note \(\because \cos{\theta}=x\)
\(\therefore \theta=\cos^{-1}{x}\)

\(2\cos^{-1}{x}=\cos^{-1}{(2x^2-1)}\)

প্রমাণঃ
ধরি,
\(\sin{\theta}=x\)
\(\Rightarrow \theta=\sin^{-1}{x}\)
আবার,
\(\sin{3\theta}=3\sin{\theta}-4\sin^3{\theta}\) ➜Note \(\because \sin{3A}=3\sin{A}-4\sin^3{A}\)

\(\Rightarrow 3\theta=\sin^{-1}{(3\sin{\theta}-4\sin^3{\theta})}\)
\(\therefore 3\sin^{-1}{x}=\sin^{-1}{(3x-4x^3)}\) ➜Note \(\because \sin{\theta}=x\)
\(\therefore \theta=\sin^{-1}{x}\)

\(3\sin^{-1}{x}=\sin^{-1}{(3x-4x^3)}\)
ধরি,
\(\cos{\theta}=x\)
\(\Rightarrow \theta=\cos^{-1}{x}\)
আবার,
\(\cos{3\theta}=4\cos^3{\theta}-3\cos{\theta}\) ➜Note \(\because \cos{3A}=4\cos^3{A}-3\cos{A}\)

\(\Rightarrow 3\theta=\cos^{-1}{(4\cos^3{\theta}-3\cos{\theta})}\)
\(\therefore 3\cos^{-1}{x}=\cos^{-1}{(4x^3-3x)}\) ➜Note \(\because \cos{\theta}=x\)
\(\therefore \theta=\cos^{-1}{x}\)

\(3\cos^{-1}{x}=\cos^{-1}{(4x^3-3x)}\)
ধরি,
\(\tan{\theta}=x\)
\(\Rightarrow \theta=\tan^{-1}{x}\)
আবার,
\(\tan{3\theta}=\frac{3\tan{\theta}-\tan^3{\theta}}{1-3\tan^2{\theta}}\) ➜Note \(\because \tan{3A}=\frac{3\tan{A}-\tan^3{A}}{1-3\tan^2{A}}\)

\(\Rightarrow 3\theta=\tan^{-1}{\left(\frac{3\tan{\theta}-\tan^3{\theta}}{1-3\tan^2{\theta}}\right)}\)
\(\therefore 3\tan^{-1}{x}=\tan^{-1}{\left(\frac{3x-x^3}{1-3x^2}\right)}\) ➜Note \(\because \tan{\theta}=x\)
\(\therefore \theta=\tan^{-1}{x}\)

\(3\tan^{-1}{x}=\tan^{-1}{\left(\frac{3x-x^3}{1-3x^2}\right)}\)
ধরি,
\(\cot{\theta}=x\)
\(\Rightarrow \theta=\cot^{-1}{x}\)
আবার,
\(\cot{3\theta}=\frac{\cot^3{\theta}-3\cot{\theta}}{3\cot^2{\theta}-1}\) ➜Note \(\because \cot{3A}=\frac{\cot^3{A}-3\cot{A}}{3\cot^2{A}-1}\)

\(\Rightarrow 3\theta=\cot^{-1}{\left(\frac{\cot^3{\theta}-3\cot{\theta}}{3\cot^2{\theta}-1}\right)}\)
\(\therefore 3\cot^{-1}{x}=\cot^{-1}{\left(\frac{x^3-3x}{3x^2-1}\right)}\) ➜Note \(\because \cot{\theta}=x\)
\(\therefore \theta=\cot^{-1}{x}\)

\(3\cot^{-1}{x}=\cot^{-1}{\left(\frac{x^3-3x}{3x^2-1}\right)}\)

প্রমাণঃ
ধরি,
\(\tan{\theta}=x\)
\(\Rightarrow \theta=\tan^{-1}{x}\)
এখন,
\(\tan{2\theta}=\frac{2\tan{\theta}}{1-\tan^2{\theta}}\)
\(\Rightarrow 2\theta=\tan^{-1}{\left(\frac{2\tan{\theta}}{1-\tan^2{\theta}}\right)}\)
\(\therefore 2\tan^{-1}{x}=\tan^{-1}{\left(\frac{2x}{1-x^2}\right)}\) ➜Note \(\because \tan{\theta}=x\)
\(\therefore \theta=\tan^{-1}{x}\)

আবার,
\(\sin{2\theta}=\frac{2\tan{\theta}}{1+\tan^2{\theta}}\)
\(\Rightarrow 2\theta=\sin^{-1}{\left(\frac{2\tan{\theta}}{1+\tan^2{\theta}}\right)}\)
\(\therefore 2\tan^{-1}{x}=\sin^{-1}{\left(\frac{2x}{1+x^2}\right)}\) ➜Note \(\because \tan{\theta}=x\)
\(\therefore \theta=\tan^{-1}{x}\)

আবার,
\(\cos{2\theta}=\frac{1-\tan^2{\theta}}{1+\tan^2{\theta}}\)
\(\Rightarrow 2\theta=\cos^{-1}{\left(\frac{1-\tan^2{\theta}}{1+\tan^2{\theta}}\right)}\)
\(\therefore 2\tan^{-1}{x}=\cos^{-1}{\left(\frac{1-x^2}{1+x^2}\right)}\) ➜Note \(\because \tan{\theta}=x\)
\(\therefore \theta=\tan^{-1}{x}\)

আবার,
\(\cot{2\theta}=\frac{\cos{2\theta}}{\sin{2\theta}}\)
\(\Rightarrow \cot{2\theta}=\frac{\frac{1-\tan^2{\theta}}{1+\tan^2{\theta}}}{\frac{2\tan{\theta}}{1+\tan^2{\theta}}}\)
\(\Rightarrow \cot{2\theta}=\frac{1-\tan^2{\theta}}{2\tan{\theta}}\)
\(\Rightarrow 2\theta=\cot^{-1}{\left(\frac{1-\tan^2{\theta}}{2\tan{\theta}}\right)}\)
\(\therefore 2\tan^{-1}{x}=\cot^{-1}{\left(\frac{1-x^2}{2x}\right)}\) ➜Note \(\because \tan{\theta}=x\)
\(\therefore \theta=\tan^{-1}{x}\)

\(\therefore 2\tan^{-1}{x}=\tan^{-1}{\left(\frac{2x}{1-x^2}\right)}=\sin^{-1}{\left(\frac{2x}{1+x^2}\right)}=\cos^{-1}{\left(\frac{1-x^2}{1+x^2}\right)}=\cot^{-1}{\left(\frac{1-x^2}{2x}\right)}\)
\(2\tan^{-1}{x}=\tan^{-1}{\left(\frac{2x}{1-x^2}\right)}=\sin^{-1}{\left(\frac{2x}{1+x^2}\right)}=\cos^{-1}{\left(\frac{1-x^2}{1+x^2}\right)}=\cot^{-1}{\left(\frac{1-x^2}{2x}\right)}\)