Ans: D
Note that $x+y+z=180^\circ$ and $x+y=90^\circ$. Then we have $z=90^\circ$ and $y=90^\circ -x$.
Note that $x+y+z=180^\circ$ and $x+y=90^\circ$. Then we have $z=90^\circ$ and $y=90^\circ -x$.
I is true.
$\begin{array}{rcl}
\text{LHS} & = & \tan x \tan y \\
& = & \tan x \tan (90^\circ – x) \\
& = & \tan x \times \dfrac{1}{\tan x} \\
& = & 1 \\
& = & \sin 90^\circ \\
& = & \sin z \\
& = & \text{RHS}
\end{array}$
II is true.
$\begin{array}{rcl}
\text{LHS} & = & \cos y +\cos z \\
& = & \cos (90^\circ – x) + \cos 90^\circ \\
& = & \sin x \\
& = & \text{RHS}
\end{array}$
III is true.
$\begin{array}{rcl}
\text{LHS} & = & \sin ^2 x +\sin ^2 y \\
& = & \sin ^2 x + \sin^2 (90^\circ – x) \\
& = & \sin ^2 x + \cos ^2 x \\
& = & 1 \\
& = & \sin^2 90^\circ \\
& = & \sin ^2 z \\
& = & \text{RHS}
\end{array}$
