- Home
- Standard 11
- Mathematics
3.Trigonometrical Ratios, Functions and Identities
hard
$\sqrt 2 + \sqrt 3 + \sqrt 4 + \sqrt 6 = . . ..$
A
$\cot 7\frac{{{1^o}}}{2}$
B
$\sin 7\frac{{{1^o}}}{2}$
C
$\sin \,{15^o}$
D
$\cos \,\,{15^o}$
(IIT-1966) (IIT-1975)
Solution
(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 $.
Standard 11
Mathematics
