Question

If in a $△ABC,△=a^2-(b-c{)}^2$, then tan A is equal to

• A. $\frac{15}{16}$
• B. $\frac{8}{15}$
• C. $\frac{8}{17}$
• D. $\frac{1}{2}$

Answer 1

Answer: (B)

$\frac{8}{15}$

Given:

$\Delta =a^2-(b-c{)}^2$

In a $\Delta ABC$:

$\Delta =\frac{1}{2}bc\sin A$

$\frac{1}{2}bc\sin A=a^2-(b-c{)}^2$

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

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

According to the Law of Cosines:

$\cos A=\frac{-(b^2+c^2-a^2)}{2bc}$

$\frac{1}{4}\sin A=-\cos A+1$

$\frac{\tan A}{4}=-1+\sqrt{1+(\tan A{)}^2}$

$1+(\tan A{)}^2=1+\frac{(\tan A{)}^2}{16}+\frac{\tan A}{2}$

$\tan A=0,\frac{8}{15}$

$\tan A\ne 0$ [since $\angle A\ne 0$]

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

Option B