A) \[\tan A-\tan B\]
B) \[\tan (A-B)\]
C) \[\tan (A+B)\]
D) \[\tan (A+2B)\]
Correct Answer: B
Solution :
\[2\tan \Alpha =3\tan B\] Þ \[\tan A=\frac{3}{2}\tan B=\frac{3}{2}t\], [Let \[\tan B=t\]] Þ \[\sin 2B=\frac{2t}{1+{{t}^{2}}},\cos 2B=\frac{1-{{t}^{2}}}{1+{{t}^{2}}}\] \ \[\frac{\left( \frac{2t}{1+{{t}^{2}}} \right)}{5-\left( \frac{1-{{t}^{2}}}{1+{{t}^{2}}} \right)}\]\[=\frac{2t}{4+6{{t}^{2}}}=\frac{t}{2+3{{t}^{2}}}=\tan (A-B)\].
You need to login to perform this action.
You will be redirected in
3 sec
