MCQ
$\sqrt 2 + \sqrt 3 + \sqrt 4 + \sqrt 6 $ is equal to
- ✓$\cot 7\frac{{{1^o}}}{2}$
- B$\sin 7\frac{{{1^o}}}{2}$
- C$\sin \,{15^o}$
- D$\cos \,\,{15^o}$
✓
Answer
Correct option: A.
$\cot 7\frac{{{1^o}}}{2}$
a
(a) We have $\cot A = \frac{{\cos A}}{{\sin A}} $
(a) 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 $.
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