Trigonometric Functions — MATHS STD 11 Science — Question

2. $\frac{\tan8\text{A}}{\tan2\text{A}}$

Solution:

We have,

$\frac{\sec8​​\text{A}-1}{\sec4\text{A}-1}=\frac{\frac{1}{\cos2\text{A}}-1}{\frac{1}{\cos}4\text{A}-1}$

$=\frac{\cos4\text{A}}{\cos8\text{A}}\times\frac{1-\cos8\text{A}}{1-\cos4\text{A}}$

$=\frac{\cos4\text{A}}{\cos8\text{A}}\times\frac{2\sin^24\text{A}}{2\sin^22\text{A}}(2\sin^2\theta-1-\cos2\theta)$

$=\frac{(2\cos4\text{A}\sin4\text{A})\sin4\text{A}}{2\times\cos8\text{A}\sin^22\text{A}}$

$=\frac{\sin8\text{A}\sin4\text{A}}{\cos8\text{A}\times\text{2}\sin2\text{A}\times\sin2\text{A}}$

$=\tan8\text{A}\times\frac{2\sin2\text{A}\times\cos2\text{A}}{2\sin2\text{A}\times\sin2\text{A}}$

$=\tan8\text{A}\times\cot2\text{A}$

$=\frac{\tan8\text{A}}{\tan2\text{A}}$