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3.Trigonometrical Ratios, Functions and Identities
hard
$\tan 7\frac{1}{2}^\circ =...$
A
$\sqrt 6 + \sqrt 3 + \sqrt 2 - 2$
B
$\sqrt 6 - \sqrt 3 + \sqrt 2 - 2$
C
$\sqrt 6 - \sqrt 3 + \sqrt 2 + 2$
D
$\sqrt 6 - \sqrt 3 - \sqrt 2 - 2$
Solution
(b) We have $\tan A = \frac{{\sin A}}{{\cos A}} $
$= \frac{{2\sin A\cos A}}{{2{{\cos }^2}A}} = \frac{{\sin 2A}}{{1 + {{\cos }^2}A}}$
Putting $A = 7{\frac{1}{2}^o} $
$\Rightarrow \tan 7{\frac{1}{2}^o} = \frac{{\sin {{15}^o}}}{{1 + \cos {{15}^o}}}$
On simplification, we get $\tan 7{\frac{1}{2}^o} = \sqrt 6 – \sqrt 3 + \sqrt 2 – 2$.
Standard 11
Mathematics
