Answer
(a) Use chain rule and trigonometric identity
(b) Using basic trigonometric differentiation formula
Work Step by Step
Step-1: Differentiate $g(x)$ using chain rule to get,
$$g(x)=\sin^2x + \cos^2x$$
$$g'(x)=2\sin x \cos x - 2 \cos x \sin x$$
$$\implies g'(x)=0$$
Step-2: Using trigonometric identity, $\sin^2 a + \cos^2 a = 1$
$$g(x) = \sin^2 x + \cos^2 x = 1$$
$$\implies g'(x) = 0$$
Step-3: By chain rule, $f'(x) = 2\sec x (\tan x \sec x)=2\sec^2 x \tan x$
And, $g'(x)=2\tan x (\sec^2 x) = 2\sec^2 x \tan x$. Thus, $$f'(x)=g'(x)$$
