Question

Evaluate : $\frac{\tan {35}^0}{\cot {55}^0}+\frac{\cos {78}^0}{\sin {12}^0}-1$

Answer 1

$\frac{\tan 35°}{\cot 55°}+\frac{\cos 78°}{\sin 12°}-1$

$=\frac{\tan (90°-55°)}{\cot 55°}+\frac{\cos (90°-12°)}{\sin 12°}-1$

$=\frac{\cot 55°}{\cot 55°}+\frac{\sin 12°}{\sin 12°}-1$

$=1+1-1$

$=1$

Hence, the answer is $1.$