Question : If $\sin \theta+\cos \theta=\frac{\sqrt{3}-1}{2 \sqrt{2}}$, then what is the value of $\tan \theta+\cot \theta$?
Option 1: $8(\sqrt{3}-2)$
Option 2: $12(\sqrt{3}-2)$
Option 3: $12(\sqrt{3}+2)$
Option 4: $8(\sqrt{3}+2)$
Correct Answer: $8(\sqrt{3}-2)$
Solution :
Given, $\sin \theta+\cos \theta=\frac{\sqrt{3}-1}{2 \sqrt{2}}$
Squaring both sides, we get,
⇒ $(\sin \theta+\cos \theta)^2=(\frac{\sqrt{3}-1}{2 \sqrt{2}})^2$
⇒ $\sin^2\theta+\cos^2\theta+2\sin\theta\cos\theta=\frac{3+1-2\sqrt3}{8}$
⇒ $1+2\sin\theta\cos\theta=\frac{4-2\sqrt{3}}{8}$
⇒ $2\sin\theta\cos\theta=\frac{2-\sqrt3}{4}-1$
⇒ $2\sin\theta\cos\theta=\frac{2-\sqrt3-4}{4}$
⇒ $\sin\theta\cos\theta=\frac{-2-\sqrt3}{8}$
Now consider, $\tan \theta+\cot \theta$
$=\frac{\sin\theta}{\cos\theta}+\frac{\cos\theta}{\sin\theta}$
$=\frac{\sin^2\theta+\cos^2\theta}{\sin\theta\cos\theta}$
$=\frac{1}{\sin\theta\cos\theta}$
$=\frac{1}{\frac{-2-\sqrt3}{8}}$
$=\frac{-8}{2+\sqrt3}$
Rationalizing it, we get,
$=\frac{-8}{2+\sqrt3}\times\frac{2-\sqrt3}{2-\sqrt3}$
$=\frac{-8(2-\sqrt3)}{2^2-(\sqrt3)^2}$
$=\frac{8(\sqrt{3}-2)}{4-3}=8(\sqrt3-2)$
Hence, the correct answer is $8(\sqrt3-2)$.
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