Calculus: Early Transcendentals 8th Edition

$y'=\dfrac{\cos t+\sin t-\sin t\sec^{2}t}{(1+\tan t)^{2}}$
$y=\dfrac{\sin t}{1+\tan t}$ Differentiate using the quotient rule: $y'=\dfrac{(1+\tan t)(\sin t)'-(\sin t)(1+\tan t)'}{(1+\tan t)^{2}}=...$ $...=\dfrac{(1+\tan t)(\cos t)-(\sin t)(\sec^{2}t)}{(1+\tan t)^{2}}=...$ Evaluate the products in the numerator: $...=\dfrac{\cos t+\cos t\tan t-\sin t\sec^{2}t}{(1+\tan t)^{2}}$ We know $\tan t=\dfrac{\sin t}{\cos t}$, so we substitute and simplify: $...=\dfrac{\cos t+(\cos t)(\dfrac{\sin t}{\cos t})-\sin t\sec^{2}t}{(1+\tan t)^{2}}=...$ $...=\dfrac{\cos t+\sin t-\sin t\sec^{2}t}{(1+\tan t)^{2}}$