Question

In $\Delta ABC$, if $2s=a+b+c$, then the value of $\frac{s(s-a)}{bc}-\frac{(s-b)(s-c)}{bc}=$

• A. $\sin A$
• B. $\cos A$
• C. $\tan A$
• D. None of these

Answer 1

Answer: (B)

$\cos A$

Now, $2s=a+b+c$, $s=\frac{a+b+c}{2}$.

Now, $\frac{s(s-a)}{bc}-\frac{(s-b)(s-c)}{bc}$

$=\frac{(a+b+c)(b+c-a)}{4bc}-\frac{(a+c-b)(a+b-c)}{4bc}$ [ Using value of $s$ we get]

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

$=\frac{b^2+c^2-a^2}{2bc}$

$=\cos A$.