Question

If $\tan A=\frac{1-\cos B}{\sin B}$, then $\tan 2A-\tan B=$

• A. $0$
• B. $1$
• C. $\frac{1}{2}$
• D. $\frac{1}{4}$

Answer 1

The correct option is A $0$

$\tan A=\frac{1-\cos B}{\sin B}$

$\tan 2A-\tan B=\left(\frac{2\tan A}{1-{\tan }^{2}A}\right)-\tan B$

$=\left[\frac{\left[\frac{2(1-\cos B)}{\sin B}\right]}{1-{\left(\frac{1-\cos B}{\sin B}\right)}^2}\right]-\tan B$

$=\left[\frac{\frac{2(1-\cos B)}{\sin B}}{\frac{{\sin }^{2}B-(1+{\cos }^{2}B-2\cos B)}{{\sin }^{2}B}}\right]-\tan B$

$=\frac{2(1-\cos B){\sin }^{2}B}{\sin B[{\sin }^{2}B-1-{\cos }^{2}B+2\cos B]}-\tan B$

$=\frac{2\sin B(1-\cos B)}{[1-{\cos }^{2}B-1-{\cos }^{2}B+2\cos B]}-\tan B$

$=\frac{2\sin B(1-\cos B)}{(2\cos B-2{\cos }^{2}B)}-\tan B$

$=\frac{2\sin B(1-\cos B)}{2\cos B(1-\cos B)}-\tan B$

$=\tan B-\tan B$

$\therefore \tan 2A-\tan B=0$