Question

The value of $\frac{2\cos {67}^o}{\sin {23}^o}-\frac{\tan {40}^o}{\cot {50}^o}$ is :

• A. $1$
• B. $0$
• C. $4$
• D. $2$

Answer 1

The correct option is A $1$

$\frac{2\cos {67}^o}{\sin {23}^o}-\frac{\tan {40}^o}{\cot {50}^o}$

$=\frac{2\cos ({90}^o-{23}^o)}{\sin {23}^o}-\frac{\tan ({90}^o-{50}^o)}{\cot {50}^o}$

$=\frac{2\sin {23}^o}{\sin {23}^o}-\frac{\cot {50}^o}{\cot {50}^o}$

$[\because \cos ({90}^o-\theta )=\sin \theta ,\tan ({90}^o-\theta )=\cot \theta ]$

$=2-1=1$