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If $\sin \theta + \cos \theta = \sqrt{3}$, then prove that $\tan \theta + \cot \theta = 1$
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Sol. $(\sin \theta + \cos \theta)^2 = (\sqrt{3})^2$ (1/2 Mark)
$\Rightarrow \sin^2 \theta + \cos^2 \theta + 2 \sin \theta \cos \theta = 3$
$\Rightarrow \sin \theta \cos \theta = 1$ --- (i) (1 Mark)
LHS $= \tan \theta + \cot \theta = \frac{\sin \theta}{\cos \theta} + \frac{\cos \theta}{\sin \theta} = \frac{\sin^2\theta+\cos^2\theta}{\cos \theta \sin \theta}$ (1 Mark)
$= \frac{1}{\cos \theta \sin \theta}$ (1/2 Mark)
$= 1$ [using (i)]
$= \text{RHS}$
$\Rightarrow \sin^2 \theta + \cos^2 \theta + 2 \sin \theta \cos \theta = 3$
$\Rightarrow \sin \theta \cos \theta = 1$ --- (i) (1 Mark)
LHS $= \tan \theta + \cot \theta = \frac{\sin \theta}{\cos \theta} + \frac{\cos \theta}{\sin \theta} = \frac{\sin^2\theta+\cos^2\theta}{\cos \theta \sin \theta}$ (1 Mark)
$= \frac{1}{\cos \theta \sin \theta}$ (1/2 Mark)
$= 1$ [using (i)]
$= \text{RHS}$