# How do you prove  sec^2(csc^2) = sec^2 + csc^2?

May 4, 2016

see below

#### Explanation:

Right Side: $= {\sec}^{2} x + {\csc}^{2} x$

$= \frac{1}{\cos} ^ 2 x + \frac{1}{\sin} ^ 2 x$

$= \frac{{\sin}^{2} x + {\cos}^{2} x}{{\cos}^{2} x {\sin}^{2} x}$

$= \frac{1}{{\cos}^{2} x {\sin}^{2} x}$

$= \frac{1}{\cos} ^ 2 x \cdot \frac{1}{\sin} ^ 2 x$

$= {\sec}^{2} x {\csc}^{2} x$

$=$ Left Side

May 4, 2016

Note the following identity:

${\csc}^{2} \theta = 1 + {\cot}^{2} \theta$

So, let's see how that works out.

$\setminus m a t h b f \left({\sec}^{2} \theta \left({\csc}^{2} \theta\right) = {\sec}^{2} \theta + {\csc}^{2} \theta\right)$

${\sec}^{2} \theta \left(1 + {\cot}^{2} \theta\right) = {\sec}^{2} \theta + {\csc}^{2} \theta$

${\sec}^{2} \theta + {\sec}^{2} \theta {\cot}^{2} \theta = {\sec}^{2} \theta + {\csc}^{2} \theta$

Lastly, use the identities

• ${\cot}^{2} \theta = {\cos}^{2} \frac{\theta}{\sin} ^ 2 \theta ,$
• $\frac{1}{\sin} ^ 2 \theta = {\csc}^{2} \theta ,$
• $\frac{1}{\cos} ^ 2 \theta = {\sec}^{2} \theta ,$

to get:

${\sec}^{2} \theta + {\sec}^{2} \theta \left({\cos}^{2} \frac{\theta}{\sin} ^ 2 \theta\right) = {\sec}^{2} \theta + {\csc}^{2} \theta$

${\sec}^{2} \theta + \frac{1}{\cancel{{\cos}^{2} \theta}} \left(\frac{\cancel{{\cos}^{2} \theta}}{\sin} ^ 2 \theta\right) = {\sec}^{2} \theta + {\csc}^{2} \theta$

${\sec}^{2} \theta + \frac{1}{\sin} ^ 2 \theta = {\sec}^{2} \theta + {\csc}^{2} \theta$

$\textcolor{b l u e}{{\sec}^{2} \theta + {\csc}^{2} \theta = {\sec}^{2} \theta + {\csc}^{2} \theta}$