Question

$(\frac{\sin {49}^{\circ }}{\cos {41}^{\circ }}-\frac{\cos {17}^{\circ }}{\sin {73}^{\circ }})=$

• A. $1$
• B. $0$
• C. $-1$
• D. None of these

Answer 1

Answer: (B)

$0$

Given

$\frac{\sin {49}^o}{\cos {41}^o}-\frac{\cos {17}^o}{\sin {73}^o}$

$=\frac{\sin {49}^o}{\cos \left({90}^o-{49}^o\right)}-\frac{\cos {17}^o}{\sin \left({90}^o-{17}^o\right)}$

$=\frac{\sin {49}^o}{\sin {49}^o}-\frac{\cos {17}^o}{\cos {17}^o}$

$=1-1=0$

0 is the correct answer