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PART - 1 CH - 3 Trigonometric Functions — MATHS STD 11 Science — Question

Rajasthan BoardEnglish MediumSTD 11 ScienceMATHSPART - 1 CH - 3 Trigonometric Functions2 Marks
Question
Prove for any triangle :
$\begin{array}{l}\tan 2 A \cdot \tan 2 B \cdot \tan 2 C=\tan 2 A+\tan 2 B +\tan 2 C\end{array}$
✓

Answer

$\because$ We know that :
$\begin{aligned}& A+B+C=180^{\circ} \\\Rightarrow \quad \therefore & 2 A+2 B+2 C=360^{\circ} \\\Rightarrow \quad & 2 A+2 B=360^{\circ}-2 C\end{aligned}$
(Taking tangent of both sides)
$\begin{array}{l}\Rightarrow \tan (2 A+2 B)=\tan \left(360^{\circ}-2 C\right) \\\Rightarrow \frac{\tan 2 A+\tan 2 B}{1-\tan 2 A \cdot \tan 2 B}=-\tan 2 C \\\qquad \quad\left[\because \tan \left(180^{\circ}-\theta\right)=-\tan \theta\right] \\\Rightarrow \tan 2 A+\tan 2 B=-\tan 2 C \cdot(1-\tan 2 A \cdot \tan 2 B) \\\Rightarrow \tan 2 A+\tan 2 B=-\tan 2 C+\tan 2 A \cdot \tan 2 B \cdot \tan 2 C \end{array}$
$\Rightarrow \tan 2 A+\tan 2 B+\tan 2 C =\tan 2 A \cdot \tan 2 B \cdot \tan 2 C$
$ \begin{array}{l} \tan 2 A \cdot \tan 2 B \cdot \tan 2 C=\tan 2 A+\tan 2 B  +\tan 2 C \text { Hence proved. } \end{array} $

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