Trigonometric Identities Self Evaluation Test Trigonometric Function Question 8
Question: If $ m=\cos \sec \theta -\sin \theta $ and $ n=\sec \theta -\cos \theta , $ then $ {m^{2/3}}+{n^{2/3}}= $
Options:
A) $ {{(mn)}^{-2/3}} $
B) $ {{(mn)}^{2/3}} $
C) $ {{(mn)}^{-1/3}} $
D) $ {{(mn)}^{1/3}} $
Answer:
Correct Answer: A
Solution:
We have, $ mn=(\cos ec,\theta -\sin \theta ),(sec\theta -\cos \theta ) $ $ =( \frac{1}{\sin \theta }-\sin \theta )( \frac{1}{\cos \theta }-\cos \theta ) $ $ =\frac{1-{{\sin }^{2}}\theta }{\sin \theta }\times \frac{1-{{\cos }^{2}}\theta }{\cos \theta } $ $ =\frac{{{\cos }^{2}}\theta }{\sin \theta }\times \frac{{{\sin }^{2}}\theta }{\cos \theta }=\sin \theta .\cos \theta $
$ \therefore {m^{2/3}}+{n^{2/3}}={{( \frac{{{\cos }^{2}}\theta }{\sin \theta } )}^{2/3}}+{{( \frac{{{\sin }^{2}}\theta }{\cos \theta } )}^{2/3}} $ $ =\frac{{{\cos }^{4/3}}\theta }{{{\sin }^{2/3}}\theta }+\frac{{{\sin }^{4/3}}\theta }{{{\cos }^{2/3}}\theta }=\frac{{{\cos }^{2}}\theta +{{\sin }^{2}}\theta }{{{(\sin \theta .\cos \theta )}^{2/3}}} $ $ =\frac{1}{{{(mn)}^{2/3}}}={{(mn)}^{-2/3}} $
BETA
