Question

sec 8 theta minus one upon sec 4 theta minus 1 is equal to tan 8 theta by tan 2 theta​

Answer 1

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Answer 2

Answer and Explanation:

To prove : $\frac{\sec 8\theta-1}{\sec 4\theta-1}=\frac{\tan8\theta}{\tan2\theta}$

Proof :

Taking LHS,

$\frac{\sec 8\theta-1}{\sec 4\theta-1}$

$=\frac{\frac{1}{\cos 8\theta}-1}{\frac{1}{\cos 4\theta}-1}$

$=\frac{(1-\cos 8\theta)\cos4\theta}{(1-\cos 4\theta)\cos8\theta}$

$=\frac{2\sin^2 4\theta\cos4\theta}{2\sin^2 2\theta\cos8\theta}$

$=\frac{2\sin 4\theta\cos4\theta\cos4\theta}{2\sin^2 2\theta\cos8\theta}$

$=\frac{\sin8\theta\cdot2\sin 2\theta\cos2\theta}{2\sin^2 2\theta\cos8\theta}$

$=\frac{\tan8\theta\cdot\cos2\theta}{\sin2\theta}$

$=\frac{\tan8\theta}{\tan2\theta}$

$=RHS$

LHS=RHS, Hence proved.