Trigonometric Identities Question 94
Question: If $ | \cos ,\theta ,{ \sin \theta +\sqrt{{{\sin }^{2}}\theta +{{\sin }^{2}}\alpha } }, |,\le k, $ then the value of k is
Options:
A) $ \sqrt{1+{{\cos }^{2}}\alpha } $
B) $ \sqrt{1+{{\sin }^{2}}\alpha } $
C) $ \sqrt{2+{{\sin }^{2}}\alpha } $
D) $ \sqrt{2+{{\cos }^{2}}\alpha } $
Show Answer
Answer:
Correct Answer: B
Solution:
Let $ u=\cos \theta { \sin \theta +\sqrt{{{\sin }^{2}}\theta +{{\sin }^{2}}\alpha } } $
Þ $ {{(u-\sin \theta \cos \theta )}^{2}}={{\cos }^{2}}\theta ({{\sin }^{2}}\theta +{{\sin }^{2}}\alpha ) $
Þ $ u^{2}{{\tan }^{2}}\theta -2u\tan \theta +u^{2}-{{\sin }^{2}}\alpha =0 $ Since tan $ \theta $ is real, therefore
Þ $ 4u^{2}-4u^{2}(u^{2}-{{\sin }^{2}}\alpha )\ge 0 $
$ \Rightarrow u^{2}-(1+{{\sin }^{2}}\alpha )\le 0 $
Þ $ |u|,\le \sqrt{1+{{\sin }^{2}}\alpha } $ .
BETA
