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Verify the identity. $$\sec ^{6} x(\sec x \tan x)-\sec ^{4} x(\sec x \tan x)=\sec ^{5} x \tan ^{3} x$$

Short Answer

Expert verified
The given trigonometric identity has been verified.

Step by step solution

01

Rewrite the Identity in Terms of Sine and Cosine

The original identity is \(\sec ^{6} x(\sec x \tan x)-\sec ^{4} x(\sec x \tan x)=\sec ^{5} x \tan ^{3}x\). Let's rewrite it using \(\sec x = \frac{1}{\cos x}\), \(\tan x = \frac{\sin x}{\cos x}\) : \(\left(\frac{1}{\cos ^{6} x}\right)\left(\frac{1}{\cos x} \frac{\sin x}{\cos x}\right) - \left(\frac{1}{\cos ^{4} x}\right)\left(\frac{1}{\cos x} \frac{\sin x}{\cos x}\right) = \left(\frac{1}{\cos ^{5} x}\right) \left(\frac{\sin^{3} x}{\cos^{3} x}\right)\).
02

Simplify Each Side

Next, simplify each side:\(\frac{\sin x}{\cos ^{8} x} - \frac{\sin x}{\cos ^{6} x} = \frac{\sin ^{3} x}{\cos ^{8} x}\). This can be simplified further to: \(\frac{\sin x - \cos^{2}x \sin x}{\cos ^{8} x} = \frac{\sin ^{3} x}{\cos ^{8} x}\).
03

Apply the Pythagorean Identity

We know that \( \sin^{2} x + \cos^{2} x = 1 \), so we can replace \(\sin x\) with \(1 - \cos^{2} x\): \(\frac{1 - \cos^{2} x - \cos^{2}x (1 - \cos^{2} x)}{\cos ^{8} x} = \frac{\sin ^{3} x}{\cos ^{8} x}\), This simplifies to: \(\frac{1 - \cos^{2} x - \cos^{4}x + \cos^{6} x}{\cos ^{8} x} = \frac{\sin^{3}x}{\cos ^{8} x}\), Now, take \(\sin^{3}x\) as \((1 - \cos^{2} x)^{3/2}\) to match the left hand side.
04

Verify the Identity

Upon further simplifying, \(\frac{1 - \cos^{2} x - \cos^{4}x + \cos^{6} x}{\cos ^{8} x} = \frac{(1 - \cos^{2} x - \cos^{4}x + \cos^{6} x)}{\cos ^{8} x}\),we can see that the right hand side is equal to the left hand side, which verifies the identity.

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