Question

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

• A. $\frac{4}{15}$
• B. $\frac{8}{15}$
• C. $\frac{11}{15}$
• D. None of these

Answer 1

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

Given, $△=a^2-{(b-c)}^2$

$\Rightarrow △=a^2-b^2-c^2+2bc\Rightarrow b^2+c^2-a^2=2bc-△$

$\Rightarrow 2bc\cos A=2bc-△[\because \cos A=\frac{b^2+c^2-a^2}{2bc}]$

$\Rightarrow 2bc\cos A=2bc-\frac{1}{2}bc\sin A[\because △=\frac{1}{2}bc\sin A]$

$\Rightarrow 4bc\cos A+bc\sin A=4bc$

$\Rightarrow 4\cos A+\sin A=4$

$\Rightarrow 4(1-2{\sin }^2\frac{A}{2})+2\sin \frac{A}{2}\cos \frac{A}{2}=4$

$\Rightarrow 2\sin \frac{A}{2}\left(\cos \frac{A}{2}-4\sin \frac{A}{2}\right)=0$

$\Rightarrow$ either $\sin \frac{A}{2}=0$ or $\cos \frac{A}{2}-4\sin \frac{A}{2}=0$

$\Rightarrow$ Either $A=0$ or $2\pi ,$which is not possible

$\Rightarrow \tan \frac{A}{2}=\frac{1}{4}.$

$\therefore \tan A=\frac{2\tan \frac{A}{2}}{1-{\tan }^2\frac{A}{2}}=\frac{2\times \frac{1}{4}}{1-\frac{1}{16}}=\frac{8}{15}.$