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  4. If θ + φ = (π/6), then (√3 + tan θ)( √3 + tan φ) =

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Q. If $\theta + \phi = \frac{\pi}{6}$, then $(\sqrt{3} + \tan \theta)( \sqrt{3} + \tan \phi) = $

Trigonometric Functions

Solution:

$\theta+\phi = \frac{\pi}{6} \Rightarrow \tan\left(\theta +\phi\right) = \tan \frac{\pi}{6} $
$ \Rightarrow \frac{\tan \theta + \tan\phi}{1- \tan\theta \tan \phi} = \frac{1}{\sqrt{3}} $
$\Rightarrow \sqrt{3} \left(\tan\theta + \tan \phi\right) = 1 - \tan \theta \tan\phi $
$ \Rightarrow \tan\theta \tan \phi + \sqrt{3} \tan\theta +\sqrt{3} \tan\phi = 1 $
Add 3 to both sides and factorise L.H.S.