Trigonometric Identities Question 77
Question: If $ x\cos \theta +y\sin \theta =z, $ then what is the value of $ {{(x\sin \theta -y\cos \theta )}^{2}} $ ?
Options:
A) $ x^{2}+y^{2}-z^{2} $
B) $ x^{2}-y^{2}-z^{2} $
C) $ x^{2}-y^{2}+z^{2} $
D) $ x^{2}+y^{2}+z^{2} $
Show Answer
Answer:
Correct Answer: A
Solution:
Here, $ z=x\cos \theta +y\sin \theta $ $ z^{2}=x^{2}{{\cos }^{2}}\theta +y^{2}{{\sin }^{2}}\theta +2xy\sin \theta \cos \theta $
$ \Rightarrow 2xy,\sin \theta \cos \theta =z^{2}-x^{2}{{\cos }^{2}}\theta -y^{2}{{\sin }^{2}}\theta $ Let, $ L={{(x\sin \theta -y\cos \theta )}^{2}} $
$ \Rightarrow L=x^{2}{{\sin }^{2}}\theta +y^{2}{{\cos }^{2}}\theta -[z^{2}-x^{2}{{\cos }^{2}}\theta -y^{2}{{\sin }^{2}}\theta ] $
$ \Rightarrow L=x^{2}[{{\sin }^{2}}\theta +{{\cos }^{2}}\theta ]+y^{2}[{{\sin }^{2}}\theta +{{\cos }^{2}}\theta ]-z^{2} $
$ \Rightarrow L=x^{2}+y^{2}-z^{2} $
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