Quantitative Aptitude Ques 2053
Question: What is $ \frac{(\sin \theta +\cos \theta )(\tan \theta +\cot \theta )}{\sec \theta +cosec\theta } $ equal to?
Options:
A) $ 1 $
B) $ 2 $
C) $ \sin \theta $
D) $ \cos \theta $
Show Answer
Answer:
Correct Answer: A
Solution:
- $ \frac{(\sin \theta +\cos \theta )(\tan \theta +\cot \theta )}{\sec \theta +cosec\theta } $
$ =\frac{(\sin \theta +\cos \theta )( \frac{\sin \theta }{\cos \theta }+\frac{\cos \theta }{\sin \theta } )}{\frac{1}{\cos \theta }+\frac{1}{\sin \theta }} $
$ =\frac{(\sin \theta +\cos \theta )( \frac{{{\sin }^{2}}\theta +{{\cos }^{2}}\theta }{\sin \theta \cos \theta } )}{\frac{\sin \theta +\cos \theta }{\sin \theta cos\theta }} $
$ =\frac{(\sin \theta +\cos \theta )( \frac{1}{\sin \theta \cos \theta } )}{\frac{\sin \theta +\cos \theta }{\sin \theta cos\theta }}=\frac{\frac{\sin \theta +\cos \theta }{\sin \theta \cos \theta }}{\frac{\sin \theta +\cos \theta }{\sin \theta \cos \theta }}=1 $
$ [\because {{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1] $
BETA
