Trigonometric Identities Question 220
Question: If $ {{\tan }^{2}}\alpha {{\tan }^{2}}\beta +{{\tan }^{2}}\beta {{\tan }^{2}}\gamma +{{\tan }^{2}}\gamma {{\tan }^{2}}\alpha $ $ +2{{\tan }^{2}}\alpha {{\tan }^{2}}\beta {{\tan }^{2}}\gamma =1, $ then the value of $ {{\sin }^{2}}\alpha +{{\sin }^{2}}\beta +{{\sin }^{2}}\gamma $ is
Options:
A) 0
B) -1
C) 1
D) None of these
Show Answer
Answer:
Correct Answer: C
Solution:
$ {{\sin }^{2}}\alpha +{{\sin }^{2}}\beta +{{\sin }^{2}}\gamma $ $ =\frac{{{\tan }^{2}}\alpha }{1+{{\tan }^{2}}\alpha }+\frac{{{\tan }^{2}}\beta }{1+{{\tan }^{2}}\beta }+\frac{{{\tan }^{2}}\gamma }{1+{{\tan }^{2}}\gamma } $ $ =\frac{x}{1+x}+\frac{y}{1+y}+\frac{z}{1+z} $ $ (x={{\tan }^{2}}\alpha ,,y={{\tan }^{2}}\beta ,,z={{\tan }^{2}}\gamma ) $ $ =\frac{(x+y+z)+(xy+yz+zx+2xyz)+xy+yz+zx+xyz}{(1+x)(1+y)(1+z)} $ $ =\frac{1+x+y+z+xy+yz+zx+xyz}{(1+x)(1+y)(1+z)}=1 $ $ (\because xy+yz+zx+2xyz=1) $
BETA
