2007-II-20

Ans: D
Since $x+y=90^\circ$, we have $x = 90^\circ – y$.

I must be true.

$\begin{array}{rcl}
\sin x & = & \sin (90^\circ-y) \\
\sin x & = & \cos y
\end{array}$

II must be true.

$\begin{array}{rcl}
\sin(90^\circ -x) & = & \sin y \\
\sin(90^\circ -x) & = & \cos (90^\circ -y)
\end{array}$

III must be true.

$\begin{array}{rcl}
\tan x \tan y & = & \tan (90^\circ – y) \tan y \\
\tan x \tan y & = & \dfrac{1}{\tan y} \times \tan y \\
\tan x \tan y & = & 1
\end{array}$