Question

Evaluate the following:

${\left(\frac{\sin 35°}{\cos 55°}\right)}^2+\left(\frac{\cos 55°}{\sin 35°}\right)-2\cos 60°$

Answer 1

Solution.

Use properties of trigonometry to find the value of given expression:

$\begin{array}{rcl}{\left(\frac{\sin 35°}{\cos 55°}\right)}^2+\left(\frac{\cos 55°}{\sin 35°}\right)-2\cos 60°& =& {\left(\frac{\sin 35°}{\cos \left(90°-35°\right)}\right)}^2+\left(\frac{\cos 55°}{\sin \left(90°-55°\right)}\right)-2\cos 60°\left[\because \sin \left(90°-\theta \right)=\cos \theta \&\cos \left(90°-\theta \right)=\sin \theta \right]\\ & =& {\left(\frac{\sin 35°}{\sin 35°}\right)}^2+\left(\frac{\cos 55°}{\cos 55°}\right)-2\times \frac{1}{2}\left(\because \cos 60°=\frac{1}{2}\right)\\ & =& 1+1-1\\ & =& 1\end{array}$

Hence, the required value is $1$.