Question

What is the value of $\left(\frac{\cos {21}^0}{\sin {69}^0}\right)-\left(\frac{\sin {40}^0}{\cos {50}^0}\right)$?

Answer 1

$\sin {69}^0=\sin ({90}^0-{21}^0)=\cos {21}^0$
and $\cos {50}^0=\cos ({90}^0-{40}^0)=\sin {40}^0$
$\therefore \frac{\cos {21}^0}{\sin {69}^0}=\frac{\cos {21}^0}{\cos {21}^0}=1$ and similarly
$\frac{\sin {40}^0}{\cos {50}^0}=1$
Hence the value of the expression
$={\left(1\right)}^2-{\left(1\right)}^2=0$