Answer
$\dfrac{(\sin t+\cos t)^{2}}{\sin t\cos t}=2+\sec t\csc t$
Work Step by Step
$\dfrac{(\sin t+\cos t)^{2}}{\sin t\cos t}=2+\sec t\csc t$
On the left side of the equation, evaluate the power in the numerator:
$\dfrac{\sin^{2}t+2\sin t\cos t+\cos^{2}t}{\sin t\cos t}=2+\sec t\csc t$
Substitute $\sin^{2}t+\cos^{2}t$ with $1$ in the numerator:
$\dfrac{1+2\sin t\cos t}{\sin t\cos t}=2+\sec t\csc t$
Rewrite the fraction on the left as $\dfrac{1}{\sin t\cos t}+\dfrac{2\sin t\cos t}{\sin t\cos t}$ and simplify:
$\dfrac{1}{\sin t\cos t}+\dfrac{2\sin t\cos t}{\sin t\cos t}=2+\sec t\csc t$
$\dfrac{1}{\sin t\cos t}+2=2+\sec t\csc t$
On the left side, $\dfrac{1}{\sin t\cos t}$ can be rewritten as $\Big(\dfrac{1}{\sin t}\Big)\Big(\dfrac{1}{\cos t}\Big)$:
$2+\Big(\dfrac{1}{\sin t}\Big)\Big(\dfrac{1}{\cos t}\Big)=2+\sec t\cos t$
Since $\dfrac{1}{\sin t}=\csc t$ and $\dfrac{1}{\cos t}=\sec t$, the identity is proved:
$2+\sec t\cos t=2+\sec t\cos t$
