Derive the formula of \(\tan(A+B)\)
\(\tan(A+B)=\frac{\sin(A+B)}{\cos(A+B)}\)
=\(\frac{\sin A\cos B+\cos A\sin B}{\cos A\cos B-\sin A\sin B}\)
Divide numerator and denominator by \(\cos A\cos B\) we get
\(\tan(A+B)=\frac{\tan A +\tan B}{1-\tan A\tan B}\)
Derive the formula of \(\tan(A-B)\)
\(\tan(A-B)=\frac{\sin(A-B)}{\cos(A-B)}\)
=\(\frac{\sin A\cos B-\cos A\sin B}{\cos A\cos B+\sin A\sin B}\)
Devide numerator and denominator by \(\cos A\cos B\) we get
\(\tan(A-B)=\frac{\tan A -\tan B}{1+\tan A\tan B}\)
Derive the formula of \(\cot(A+B)\)
\(\cot(A+B)=\frac{\cos(A+B)}{\sin(A+B)}\)
=\(\frac{\cos A\cos B-\sin A\sin B}{\sin A\cos B+\cos A\sin B}\)
Divide numerator and denominator by \(\sin A\sin B\) we get
\(\cot(A+B)=\frac{\cot A \cot B -1}{\cot B+\cot A}\)
Derive the formula of \(\cot(A-B)\)
\(\cot(A-B)=\frac{\cos(A-B)}{\sin(A-B)}\)
=\(\frac{\cos A\cos B+\sin A\sin B}{\sin A\cos B-\cos A\sin B}\)
Divide numerator and denominator by \(\sin A \sin B\) we get
\(\cot(A-B)=\frac{\cot A \cot B -1}{\cot B-\cot B}\)
Derive the formula of \(\tan(A-B)\)
\(\tan(A-B)=\frac{\sin(A-B)}{\cos(A-B)}\)
=\(\frac{\sin A\cos B-\cos A\sin B}{\cos A\cos B+\sin A\sin B}\)
Divide numerator and denominator by \(\cos A\cos B\) we get
\(\tan(A+B+C)\)=\(\frac{\tan A +\tan (B+C)}{1-\tan A\tan (B+C)}\)
=\(\frac{\tan A +\frac{\tan B +\tan C}{1-\tan B\tan C}}{1-\tan A \times \frac{\tan B +\tan C}{1-\tan B \tan C}}\)
=\(\frac{\tan A -\tan A \tan B \tan C +\tan B + \tan C}{1-\tan B \tan C-\tan C \tan A-\tan A \tan B}\)
Formula:
\(\tan(A+B)=\frac{\tan A +\tan B}{1-\tan A\tan B}\)
\(\tan(A-B)=\frac{\tan A -\tan B}{1+\tan A\tan B}\)
\(\cot(A+B)=\frac{\cot A \cot B -1}{\cot B+\cot A}\)
\(\cot(A-B)=\frac{\cot A \cot B -1}{\cot B-\cot B}\)
\(\tan(A+B+C)=\frac{\tan A -\tan A \tan B \tan C +\tan B + \tan C}{1-\tan B \tan C-\tan C \tan A-\tan A \tan B}\)