Question

The value of $\frac{\cos {58}^{\circ }}{\sin {32}^{\circ }}+\frac{\sin {22}^{\circ }}{\cos {68}^{\circ }}-\frac{\cos {38}^{\circ }\text{cosec}{52}^{\circ }}{\tan {18}^{\circ }\tan {35}^{\circ }\tan {72}^{\circ }\tan {55}^{\circ }}$
is

• A. $0$
• B. $\frac{1}{2}$
• C. $1$
• D. $2$

Answer 1

The correct option is C $1$

$\frac{\cos {58}^{\circ }}{\sin {32}^{\circ }}+\frac{\sin {22}^{\circ }}{\cos {68}^{\circ }}-\frac{\cos {38}^{\circ }\text{cosec}{52}^{\circ }}{\tan {18}^{\circ }\tan {35}^{\circ }\tan {72}^{\circ }\tan {55}^{\circ }}=\frac{\cos ({90}^{\circ }-{32}^{\circ })}{\sin {32}^{\circ }}+\frac{\sin ({90}^{\circ }-{68}^{\circ })}{\cos {68}^{\circ }}-\frac{\cos {38}^{\circ }\text{cosec}{52}^{\circ }}{\tan {18}^{\circ }\tan {35}^{\circ }\tan {72}^{\circ }\tan {55}^{\circ }}=1+1-\frac{\cos {38}^{\circ }\text{cosec}({90}^{\circ }-{38}^{\circ })}{\tan {18}^{\circ }\tan {35}^{\circ }\tan {72}^{\circ }\tan {55}^{\circ }}=2-\frac{1}{\tan {18}^{\circ }\tan {35}^{\circ }\tan {72}^{\circ }\tan {55}^{\circ }}$

Now,
$\tan {18}^{\circ }\tan {35}^{\circ }\tan {72}^{\circ }\tan {55}^{\circ }=\tan {18}^{\circ }\tan {35}^{\circ }\tan ({90}^{\circ }-{18}^{\circ })\tan ({90}^{\circ }-{35}^{\circ })=\tan {18}^{\circ }\tan {35}^{\circ }\cot {18}^{\circ }\cot {35}^{\circ }=\left(\tan {18}^{\circ }\cot {18}^{\circ }\right)\left(\tan {35}^{\circ }\cot {35}^{\circ }\right)=1$

Therefore,
$\frac{\cos {58}^{\circ }}{\sin {32}^{\circ }}+\frac{\sin {22}^{\circ }}{\cos {68}^{\circ }}-\frac{\cos {38}^{\circ }\text{cosec}{52}^{\circ }}{\tan {18}^{\circ }\tan {35}^{\circ }\tan {72}^{\circ }\tan {55}^{\circ }}=2-1=1$