Explanation

(a)
Given, $y = (1 + x)^{y} + \sin^{- 1}\left( \sin^{2}x \right)$
At $x = 0,\ \ y = 1$
Let $y = u + v$ where $u = (1 + x)^{y},\ v = \sin^{- 1}{(\sin^{2}x)}$
$\Rightarrow \ \ \ \ \log{u = y\log(1 + x)}$
$\Rightarrow \ \ \ \ \ \frac{du}{dx}\frac{1}{u} = \frac{y}{1 + x} + \log{(1 + x)\frac{dy}{dx}}$
and $\frac{dv}{dx} = \frac{1}{\sqrt{1 - \sin^{2}x}}.2\sin{x\cos x}$
$\therefore\ \ \frac{dy}{dx} = (1 + x)^{y}\ \left\lbrack \frac{y}{1 + x} + \log{(1 + x)\frac{dy}{dx}\ } \right\rbrack + \frac{\sin{2x}}{\cos x}$
$\Rightarrow \ \ \ \ \frac{dy}{dx}\left\lbrack 1 - (1 + x)^{y}\log(x + 1) \right\rbrack = \frac{(1 + x)^{y}y}{1 + x} + \frac{\sin{2x}}{\cos x}$
$\therefore\ \ \ \left. \ \frac{dy}{dx} \right|_{(0,\ 1)} = \frac{1}{1 + 0} = 1$
$\therefore$ Equation of tangent is
$(y - 1) = 1(x)\ \ \Rightarrow \ \ \ \ \ x - y + 1 = 0$