Answer
$t^{\prime}(x)=\sin x (1+\sec x)+\sec x \tan x (1-\cos x)$
Work Step by Step
We have: $t(x)=(1+\sec x)(1-\cos x)$
We need to differentiate both sides with respect to $x$.
$t^{\prime}(x)=\dfrac{d}{dx} [(1+\sec x)(1-\cos x)]\\=(1+\sec x) \times (-1)(-\sin x )+(1-\cos x) \sec x \tan x\\=\sin x (1+\sec x)+\sec x \tan x (1-\cos x)$
Therefore, $t^{\prime}(x)=\sin x (1+\sec x)+\sec x \tan x (1-\cos x)$
