| What is\[\frac{(\sin \theta +\cos \theta )(\tan \theta +\cot \theta )}{\sec \theta +\text{cosec}\theta }\]equal to? |
A) \[1\]
B) \[2\]
C) \[\sin \theta \]
D) \[\cos \theta \]
Correct Answer: A
Solution :
| \[\frac{(\sin \theta +\cos \theta )(\tan \theta +\cot \theta )}{\sec \theta +\text{cosec}\theta }\] |
| \[=\frac{(\sin \theta +\cos \theta )\left( \frac{\sin \theta }{\cos \theta }+\frac{\cos \theta }{\sin \theta } \right)}{\frac{1}{\cos \theta }+\frac{1}{\sin \theta }}\] |
| \[=\frac{(\sin \theta +\cos \theta )\left( \frac{{{\sin }^{2}}\theta +{{\cos }^{2}}\theta }{\sin \theta \cos \theta } \right)}{\frac{\sin \theta +\cos \theta }{\sin \theta cos\theta }}\] |
| \[=\frac{(\sin \theta +\cos \theta )\left( \frac{1}{\sin \theta \cos \theta } \right)}{\frac{\sin \theta +\cos \theta }{\sin \theta cos\theta }}=\frac{\frac{\sin \theta +\cos \theta }{\sin \theta \cos \theta }}{\frac{\sin \theta +\cos \theta }{\sin \theta \cos \theta }}=1\] |
| \[[\because {{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1]\] |
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