# Verify the identity of sec^6X(secXtanX)-sec^4X(secXtanX)=sec^5Xtan^3X ?

Apr 12, 2018

See below

#### Explanation:

${\sec}^{6} x \left(\sec x \tan x\right) - {\sec}^{4} x \left(\sec x \tan x\right) = {\sec}^{4} x \sec x \tan x \left({\sec}^{2} x - 1\right) = {\sec}^{5} x \tan x {\tan}^{2} x = {\sec}^{5} x {\tan}^{3} x$

Apr 12, 2018

${\sec}^{6} X \left(\sec X \tan X\right) - {\sec}^{4} X \left(\sec X \tan X\right) = {\sec}^{5} X {\tan}^{3} X$

Taking the $L H S$,

${\sec}^{6} X \left(\sec X \tan X\right) - {\sec}^{4} X \left(\sec X \tan X\right)$

$\implies \left({\sec}^{6} X - {\sec}^{4} X\right) \left(\sec X \tan X\right)$

$\implies {\sec}^{4} X \left({\sec}^{2} X - 1\right) \left(\sec X \tan X\right)$

$\implies {\sec}^{4} X \left({\tan}^{2} X\right) \left(\sec X \tan X\right)$ color(white)(ww $\left[\text{as } \textcolor{red}{{\sec}^{2} X - 1 = {\tan}^{2} X}\right]$

Getting the $\tan X$ and $\sec X$ together,

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

Hence Verified ! :)

Apr 12, 2018

#### Explanation:

We know that,

color(red)((1)sec^2theta-1=tan^2theta

Here,

${\sec}^{6} X \left(\sec X \tan X\right) - {\sec}^{4} X \left(\sec X \tan X\right) = {\sec}^{5} X {\tan}^{3} X$

We take,

$L H S = {\sec}^{6} X \left(\textcolor{b l u e}{\sec X \tan X}\right) - {\sec}^{4} X \left(\textcolor{b l u e}{\sec X \tan X}\right)$

$= \textcolor{b l u e}{\sec X \tan X} \left({\sec}^{6} X - {\sec}^{4} X\right)$

$= \textcolor{b l u e}{\sec X \tan X} {\sec}^{4} X \left({\sec}^{2} X - 1\right)$

$= {\sec}^{5} X \tan X \left(\textcolor{red}{{\sec}^{2} X - 1}\right) \ldots \to A p p l y \left(1\right)$

$= {\sec}^{5} X \tan X \left(\textcolor{red}{{\tan}^{2} X}\right)$

$= {\sec}^{5} X {\tan}^{3} X$

$= R H S$