Trigonometric Identities Question 227
Question: The value of $ \frac{\cot 54{}^\circ }{\tan 36{}^\circ }+\frac{\tan 20{}^\circ }{\cot 70{}^\circ } $ is
[Karnataka CET 1999]
Options:
A) 2
B) 3
C) 1
D) 0
Show Answer
Answer:
Correct Answer: A
Solution:
$ \tan (90{}^\circ -\theta )=\cot \theta ,\ \cot (90{}^\circ -\theta )=\tan \theta . $ Therefore $ \frac{\cot 54{}^\circ }{\tan 36{}^\circ }+\frac{\tan 20{}^\circ }{\cot 70{}^\circ } $ $ =\frac{\cot 54{}^\circ }{\tan (90{}^\circ -54{}^\circ )}+\frac{\tan 20{}^\circ }{(\cot 90{}^\circ -20{}^\circ )} $ $ \frac{\cot 54{}^\circ }{\cot 54{}^\circ }+\frac{\tan 20{}^\circ }{\tan 20{}^\circ }=1+1=2 $ .
BETA
