Question

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

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

Answer 1

The correct option is C

$\frac{8}{15}$

Step 1. Find the value of $\mathit{t}\mathit{a}\mathit{n}\mathbf{}\mathit{A}$:

Given, $∆=a^2-{(b-c)}^2$ …..(1)

As we know,

$s=\frac{(a+b+c)}{2}$

$\Rightarrow$$a=2s–(b+c)$

Step 2. Put the value of $a$ in equation (1), we get

$\begin{array}{rcl}∆& =& {[2s–(b+c\left)\right]}^2–(b–c{)}^2\\ & =& [4s^2+{(b+c)}^2–4s(b+c\left)\right]–{(b–c)}^2=4s^2–4s(b+c)+[{(b+c)}^2–{(b–c)}^2]=4s^2–4s(b+c)+4bc\\ & =& 4s^2–4bs–4cs+4bc\\ & =& 4s(s–b)–4c(s–b)\\ & =& (4s–4c)(s-b)\\ & =& 4(s–c)(s–b)\end{array}$

$\frac{1}{4}=\frac{(s–c)(s–b)}{∆}$ …(2)

As We know,

$\tan \left(\frac{A}{2}\right)=\sqrt{\left[\frac{(s–b)(s–c)}{s(s–a)}\right]}$

$\Rightarrow$$\sqrt{\left[(s–b)(s–c)\right]}=\sqrt{\left[s(s–a)\right]}\tan \left(\frac{A}{2}\right)$

Step 3. On Multiply both sides by $\sqrt{\left[(s–b)(s–c)\right]}$, we get

$(s–b)(s–c)=\sqrt{\left(s\right(s–a\left)\right(s–b\left)\right(s–c\left)\right)}\tan \left(\frac{A}{2}\right)$

$\Rightarrow$ $(s–b)(s–c)=\tan \left(\frac{A}{2}\right)∆$ $\left[\mathbf{\because }\mathbf{△}\mathbf{}\mathbf{=}\sqrt{\mathbf{s}\mathbf{(}\mathbf{s}\mathbf{}\mathbf{-}\mathbf{a}\mathbf{)}\mathbf{(}\mathbf{s}\mathbf{}\mathbf{-}\mathbf{}\mathbf{b}\mathbf{)}\mathbf{(}\mathbf{s}\mathbf{}\mathbf{-}\mathbf{c}\mathbf{)}}\right]$

$\Rightarrow$$\frac{(s–b)(s–c)}{∆}=\tan \left(\frac{A}{2}\right)$ …(3)

Step 4. By comparing equation (2) and (3), we get

$\tan \left(\frac{A}{2}\right)=\frac{1}{4}$

As we know,

$\begin{array}{rcl}\mathit{t}\mathit{a}\mathit{n}\mathbf{}\mathit{A}\mathbf{}& \mathbf{=}& \mathbf{}\frac{\mathbf{2}\mathbf{}\mathbf{t}\mathbf{a}\mathbf{n}\mathbf{}\mathbf{\left(}{\displaystyle \frac{A}{2}}\mathbf{\right)}}{\mathbf{1}\mathbf{}\mathbf{–}\mathbf{}\mathbf{t}\mathbf{a}{\mathbf{n}}^{\mathbf{2}}\mathbf{\left(}{\displaystyle \frac{A}{2}}\mathbf{\right)}}\\ & =& \frac{2\times {\displaystyle \frac{1}{4}}}{1–{\displaystyle \frac{1}{16}}}\\ & =& \frac{8}{15}\end{array}$

$\mathbf{\therefore }\mathit{t}\mathit{a}\mathit{n}\mathbf{}\mathit{A}\mathbf{}\mathbf{=}\mathbf{}\frac{\mathbf{8}}{\mathbf{15}}$

Hence, Option ‘C’ is Correct.