Answer
$\tan t+cosec t \sec t=\dfrac{\sin^2 t+1}{\sin t\cos t}$
Work Step by Step
Consider left hand side :
$\tan t+cosec t \sec t=\dfrac{\sin t}{\cos t}+\dfrac{1}{\sin t} \cdot \dfrac{1}{\cos t}$
or, $\dfrac{\sin t}{\cos t}+\dfrac{1}{\sin t \cos t}=\dfrac{\sin^2 t+1}{\sin t\cos t}$ (RHS)
Thus, $\tan t+cosec t \sec t=\dfrac{\sin^2 t+1}{\sin t\cos t}$
Hence, the left-hand side and right hand side are equal.
