# Prove that sec^4-cos^4=1-2cos^2 ?

Jul 3, 2018

$\textcolor{red}{\text{The question is wrong!!!!}}$

#### Explanation:

$\textcolor{b l u e}{\text{Approaching the question on the basis of the information that you have provided, The solution would be:-}}$

${\sec}^{4} x - {\cos}^{4}$

= $\left({\sec}^{2} x + {\cos}^{2} x\right) \left({\sec}^{2} x - {\cos}^{2} x\right)$

= $\frac{\left(1 + {\cos}^{4} x\right) \left(1 - {\cos}^{4} x\right)}{\cos} ^ 4 x$

= $\left(1 + {\cos}^{4} x\right) \left(1 - {\cos}^{2} x\right) \frac{1 + {\cos}^{2} x}{\cos} ^ 4 x$

$\textcolor{b l u e}{\text{Thus the solution would be not coming equal to RHS}}$.

The correct question would be ${\sin}^{4} x$ in place of ${\sec}^{4} x$.

On solving this question we get,

${\sin}^{4} x - {\cos}^{4} x = 1 - 2 {\cos}^{2} x ,$ then

LHS = ${\sin}^{4} x - {\cos}^{4} x$

= $\left({\sin}^{2} x + {\cos}^{2} x\right) \left({\sin}^{2} x - {\cos}^{2} x\right)$

= $1 \cdot \left({\sin}^{2} x - {\cos}^{2} x\right)$

= $1 - {\cos}^{2} x - {\cos}^{2} x$

= $1 - 2 {\cos}^{2} x$ = RHS