If $\alpha, \beta$ and $\gamma$ are length of the altitudes of a $\triangle A B C$ with area $\Delta$, then $\frac{\Delta^2}{R^2}\left(\frac{1}{\alpha^2}+\frac{1}{\beta^2}+\frac{1}{\gamma^2}\right)$ is equal to

• Correct answer
  A

  $\sin ^2 A+\sin ^2 B+\sin ^2 C$
• B

  $\cos ^2 A+\cos ^2 B+\cos ^2 C$
• C

  $\tan ^2 A+\tan ^2 B+\tan ^2 C$
• D

  $\cot ^2 A+\cot ^2 B+\cot ^2 C$

Solution

$\because$ Area of triangle $=\frac{1}{2}$ base $\times$ altitude $\therefore \Delta=\frac{1}{2} a \alpha=\frac{1}{2} b \beta=\frac{1}{2} c \gamma$
$\Rightarrow \alpha=\frac{2 \Delta}{a}, \beta=\frac{2 \Delta}{b}$ and $\gamma=\frac{2 \Delta}{c}$ $\therefore \frac{\Delta^2}{R^2}\left(\frac{1}{\alpha^2}+\frac{1}{\beta^2}+\frac{1}{\gamma^2}\right)$ $=\frac{\Delta^2}{R^2}\left(\frac{a^2}{4 \Delta^2}+\frac{b^2}{4 \Delta^2}+\frac{c^2}{4 \Delta^2}\right)$ $\begin{aligned} & =\frac{1}{4 R^2}\left(4 R^2 \sin ^2 A+4 R^2 \sin ^2 B+4 R^2 \sin ^2 C\right) \\ & \qquad\left[\because \frac{\sin A}{a}=\frac{\sin B}{b}=\frac{\sin C}{c}=\frac{1}{2 R}\right] \\ & =\sin ^2 A+\sin ^2 B+\sin ^2 C\end{aligned}$

Hence, the correct answer is option: 'A'.