# How do you prove sec^6x(secxtanx) - sec^4x(secxtanx) = sec^5xtan^3x?

Mar 11, 2018

$L H S = {\sec}^{6} x \left(\sec x \tan x\right) - {\sec}^{4} x \left(\sec x \tan x\right)$

$= {\sec}^{5} x \tan x \left({\sec}^{2} x - 1\right)$

$= {\sec}^{5} x \tan x \cdot {\tan}^{2} x$

$= {\sec}^{5} x {\tan}^{3} x = R H S$

Mar 11, 2018

As proved.

#### Explanation:

To prove ${\sec}^{6} x \left(\sec s \tan x\right) - {\sec}^{4} x \left(\sec x \tan x\right)$ = sec^5x tan^3x

$L H S = {\sec}^{6} x \left(\sec x \tan x\right) - {\sec}^{4} x \left(\sec x \tan x\right)$

=> sec x tan x) ( sec ^6x - sec ^4x) taking common term color(brown)(sec x tan x out.

$\implies \left(\sec x \tan x\right) \cdot \left({\sec}^{4} x\right) \left({\sec}^{2} x - 1\right)$ taking color(brown)(sec^4x# out. $\implies \left(\sec x \tan x\right) \left({\sec}^{4} x\right) \left({\tan}^{2} x\right)$ as ${\sec}^{2} x - 1 = {\tan}^{2} x$

$\implies \left(\sec x \cdot {\sec}^{4} x\right) \cdot \left(\tan x \cdot {\tan}^{2} x\right)$ Rearranging like terms together.

$\implies {\sec}^{5} x \cdot {\tan}^{3} x = R H S$

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