Answer
$\dfrac{\sec x+\csc x}{\tan x+\cot x}=\sin x+\cos x$
Work Step by Step
$\dfrac{\sec x+\csc x}{\tan x+\cot x}=\sin x+\cos x$
On the left side of the equation, replace $\sec x$ by $\dfrac{1}{\cos x}$, $\csc x$ by $\dfrac{1}{\sin x}$, $\tan x$ by $\dfrac{\sin x}{\cos x}$ and $\cot x$ by $\dfrac{\cos x}{\sin x}$:
$\dfrac{\dfrac{1}{\cos x}+\dfrac{1}{\sin x}}{\dfrac{\sin x}{\cos x}+\dfrac{\cos x}{\sin x}}=\sin x+\cos x$
Evaluate the sums indicated on the left side:
$\dfrac{\dfrac{\sin x+\cos x}{\sin x\cos x}}{\dfrac{\sin^{2}x+\cos^{2}x}{\sin x\cos x}}=\sin x+\cos x$
Evaluate the division on the left side:
$\dfrac{\sin x+\cos x}{\sin^{2}x+\cos^{2}x}=\sin x+\cos x$
Since $\sin^{2}x+\cos^{2}x=1$, the identity is proved:
$\sin x+\cos x=\sin x+\cos x$
