Trigonometric Identities Question 192
Question: If $ {{\tan }^{2}}\frac{\pi -A}{4}+{{\tan }^{2}}\frac{\pi -B}{4}+{{\tan }^{2}}\frac{\pi -C}{4}=1,then\Delta ABCis $
Options:
A) equilateral triangle
B) isosceles triangle
C) scalene
D) none of these
Show Answer
Answer:
Correct Answer: A
Solution:
[a] Let $ \alpha =\frac{\pi -A}{4},\beta =\frac{\pi -B}{4},\gamma =\frac{\pi -C}{4} $
$ \Rightarrow ,\alpha +\beta +\gamma =\frac{\pi }{2} $
$ \Rightarrow ,\tan \alpha \tan \beta =1 $
$ \Rightarrow ,\sum{{{\tan }^{2}}\alpha =1=\sum{\tan \alpha \tan \beta } $
$ \Rightarrow $ $ \tan \alpha =\tan \beta =\tan \gamma $
BETA
