Question : If $\sin \alpha+\cos \alpha=\frac{2}{\sqrt{3}}$, then what is $(\tan \alpha+\cot \alpha)$ equal to?
Option 1: 1
Option 2: 2
Option 3: 6
Option 4: 0
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Correct Answer: 6
Solution :
$\sin \alpha+\cos \alpha=\frac{2}{\sqrt{3}}$
Square both sides,
$(\sin \alpha+\cos \alpha)^2=\left(\frac{2}{\sqrt{3}}\right)^2$
⇒ $\sin^2 \alpha + 2\sin \alpha \cos \alpha + \cos^2 \alpha = \frac{4}{3}$
⇒ $1 + 2\sin \alpha \cos \alpha = \frac{4}{3}$
⇒ $2\sin \alpha \cos \alpha = \frac{4}{3} - 1 = \frac{1}{3}$
⇒ $\sin \alpha \cos \alpha = \frac{1}{6}$ ...........................................(i)
Now,
$\tan \alpha + \cot \alpha = \frac{\sin \alpha}{\cos \alpha} + \frac{\cos \alpha}{\sin \alpha} = \frac{\sin^2 \alpha + \cos^2 \alpha}{\sin \alpha \cos \alpha}$
From (i)
$\tan \alpha + \cot \alpha = \frac{1}{\frac{1}{6}} = 6$
So, $\tan \alpha + \cot \alpha = 6$
Hence, the correct answer is 6.
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