Question

In $\Delta$ $ABC$ the sides opposite to angles $A,B,C$ are denoted by $a,b,c$ respectively.

If the area of a triangle $ABC$ is given by $\Delta =a^2-(b-c{)}^2$, then $\tan A$ is equal to?

• A. $\frac{1}{4}$
• B. $\frac{8}{15}$
• C. $\frac{4}{15}$
• D. $\frac{3}{4}$

Answer 1

The correct option is B $\frac{8}{15}$

$△=a^2-{(b-c)}^2\Rightarrow a^2-(b^2+c^2-2bc)△=a^2-b^2-c^2+2bc=-(b^2+c^2-a^2)+2bc=-(2bc\cos A)+2bc=2bc(1-\cos A)△=\frac{1}{2}bc\sin A=2bc(1-\cos A)\frac{\sin A}{1-\cos A}=4$

$\frac{2\sin \frac{A}{2}\cos \frac{A}{2}}{2\sin {}^2\frac{A}{2}}=4\tan \frac{A}{2}=\frac{1}{4}\tan A=\frac{2\tan \frac{A}{2}}{1-\tan {}^2\frac{A}{2}}=\frac{2\times \frac{1}{4}}{1-{\left(\frac{1}{4}\right)}^2}=\frac{\frac{1}{2}}{\frac{15}{16}}$

$\tan A=\frac{8}{15}$

(B) is correct