- Home
- Standard 11
- Mathematics
3.Trigonometrical Ratios, Functions and Identities
easy
$\frac{{\sec 8A - 1}}{{\sec 4A - 1}} = $
A
$\frac{{\tan 2A}}{{\tan 8A}}$
B
$\frac{{\tan 8A}}{{\tan 2A}}$
C
$\frac{{\cot 8A}}{{\cot 2A}}$
D
એકપણ નહિ.
Solution
(b) $\frac{{\sec 8A – 1}}{{\sec 4A – 1}}$
$ = \frac{{1 – \cos 8A}}{{\cos 8A}}.\frac{{\cos 4A}}{{1 – \cos 4A}}$
$ = \frac{{2{{\sin }^2}4A}}{{\cos 8A}}\frac{{\cos 4A}}{{2{{\sin }^2}2A}}$
$ = \frac{{2\sin 4A\cos 4A}}{{\cos 8A}}\frac{{\sin 4A}}{{2{{\sin }^2}2A}}$
$ = \tan 8A\frac{{2\sin 2A\cos 2A}}{{2{{\sin }^2}2A}} $
$= \frac{{\tan 8A}}{{\tan 2A}}.$
Standard 11
Mathematics
