Trigonometric Identities Question 229
Question: If $ \tan x=\frac{2b}{a-c}(a\ne c), $ $ y=a,{{\cos }^{2}}x+2b,\sin x\cos x+c,{{\sin }^{2}}x $ and $ z=a{{\sin }^{2}}x-2b\sin x\cos x+c{{\cos }^{2}}x, $ then
Options:
A) $ y=z $
B) $ y+z=a+c $
C) $ y-z=a+c $
D) $ y-z={{(a-c)}^{2}}+4b^{2} $
Show Answer
Answer:
Correct Answer: B
Solution:
We have, $ y+z=a({{\cos }^{2}}x+{{\sin }^{2}}x)+c({{\sin }^{2}}x+{{\cos }^{2}}x)=a+c $ $ (\therefore \text{solution is (b)}} $ $ y-z=a({{\cos }^{2}}x-{{\sin }^{2}}x)+4b\sin x\cos x $ $ -c({{\cos }^{2}}x-{{\sin }^{2}}x) $ $ =(a-c)\cos 2x+2b\sin 2x $ $ =(a-c).,( \frac{1-{{\tan }^{2}}x}{1+{{\tan }^{2}}x} )+2b.( \frac{2\tan x}{1+{{\tan }^{2}}x} ) $ $ =(a-c).{ \frac{1-4b^{2}/{{(a-c)}^{2}}}{1+4b^{2}/{{(a-c)}^{2}}} }+2b.{ \frac{2.2b/(a-c)}{1+4b^{2}/{{(a-c)}^{2}}} } $ Since $ \tan x=\frac{2b}{(a-c)} $ ,
$ \therefore y-z=\frac{(a-c).{{{(a-c)}^{2}}-4b^{2}}+8b^{2}(a-c)}{{{(a-c)}^{2}}+4b^{2}} $ $ =\frac{(a-c){{(a-c)}^{2}}+4b^{2}}{{{{(a-c)}^{2}}+4b^{2}}}=(a-c) $
$ \Rightarrow y\ne z,,,(\because a\ne c) $ .
BETA
