8A - 1 4A - 1 = - Vidyadip

MCQ

$\frac{{\sec 8A - 1}}{{\sec 4A - 1}} = $

A
$\frac{{\tan 2A}}{{\tan 8A}}$
✓
$\frac{{\tan 8A}}{{\tan 2A}}$
C
$\frac{{\cot 8A}}{{\cot 2A}}$
D
None of these

✓

Answer

Correct option: B.

$\frac{{\tan 8A}}{{\tan 2A}}$

b
(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}} $

$= \frac{{\tan 8A}}{{\tan 2A}}.$

Need a full question paper?

Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.

Start Generating Free

Explore more

More "MCQ" from 3. trignometrical ratios,functions and identities→3. trignometrical ratios,functions and identities — all question types→MATHS STD 11 Science question papers→Gujarat Board STD 11 Science question papers→Gujarat Board question paper generator→

Similar questions

The period of the function $\left| {\,\sin \pi \,x\,} \right| $ is

Three numbers are in $A.P.$ whose sum is $33$ and product is $792$, then the smallest number from these numbers is

The value of $(1 - \omega + {\omega ^2})\,{(1 - {\omega ^2} + \omega )^6}$, where $\omega ,{\omega ^2}$ are cube roots of unity

If the coefficient of the middle term in the expansion of ${(1 + x)^{2n + 2}}$ is $p$ and the coefficients of middle terms in the expansion of ${(1 + x)^{2n + 1}}$ are $q$ and $r$, then

The length of the latus rectum of the parabola $9{x^2} - 6x + 36y + 19 = 0$

The ratio in which $x$-axis divides the join of the points $(2, -3)$ and $(5, 6)$ is

If $\frac{{1 - {{\tan }^2}\theta }}{{{{\sec }^2}\theta }} = \frac{1}{2}$, then the general value of $\theta $ is

Let $p$ and $q$ be two real numbers such that $p+q=$ 3 and $p^{4}+q^{4}=369$. Then $\left(\frac{1}{p}+\frac{1}{q}\right)^{-2}$ is equal to

A variable line $\mathrm{L}$ passes through the point $(3,5)$ and intersects the positive coordinate axes at the points $\mathrm{A}$ and $\mathrm{B}$. The minimum area of the triangle $\mathrm{OAB}$, where $\mathrm{O}$ is the origin, is :

If P(n) = 2 + 4 + ......+ 2n, n ϵ N, then P(k) = k(k + 1) + 2 ⇒ P(k) = k(k + 1) + 2 for all k ϵ N. S we can conclude that P(n) = n(n + 1) + 2 for