Trigonometric Identities Question 317
Question: $ \sqrt{2}+\sqrt{3}+\sqrt{4}+\sqrt{6} $ is equal to
[IIT 1966, 1975]
Options:
A) $ \cot 7\frac{1^{o}}{2} $
B) $ \sin 7\frac{1^{o}}{2} $
C) $ \sin ,15^{o} $
D) $ \cos 15^{o} $
Show Answer
Answer:
Correct Answer: A
Solution:
We have $ \cot A=\frac{\cos A}{\sin A}=\frac{2{{\cos }^{2}}A}{2\sin A\cos A}=\frac{1+\cos 2A}{\sin 2A} $ Putting $ A=7\frac{1^{o}}{2}\Rightarrow \cot 7\frac{1^{o}}{2}=\frac{1+\cos 15^{o}}{\sin 15^{o}} $ On simplification, we get $ \cot 7\frac{1^{o}}{2}=\sqrt{6}+\sqrt{2}+\sqrt{3}+\sqrt{4} $ .
BETA
