Trigonometric Identities Question 14
Question: If $ \tan \alpha $ equals the integral solution of the inequality $ 4x^{2}-16x+15<0 $ and $ \cos \beta $ equals to the slope of the bisector of first quadrant, then $ \sin (\alpha +\beta )\sin (\alpha -\beta ) $ is equal to
[Kerala (Engg.) 1993]
Options:
A) $ \frac{3}{5} $
B) $ -\frac{3}{5} $
C) $ \frac{2}{\sqrt{5}} $
D) $ \frac{4}{5} $
Show Answer
Answer:
Correct Answer: D
Solution:
We have $ 4x^{2}-16x+15<0\Rightarrow ,\frac{3}{2}<x<\frac{5}{2} $
$ \therefore $ Integral solution of $ 4x^{2}-16x+15<0 $ is x = 2. Thus $ \tan \alpha =2. $ It is given that $ \cos \beta =\tan 45^{o}=1 $
$ \therefore \sin ,(\alpha +\beta ),\sin ,(\alpha -\beta )={{\sin }^{2}}\alpha -{{\sin }^{2}}\beta $ $ =\frac{1}{1+{{\cot }^{2}}\alpha }-(1-{{\cos }^{2}}\beta )=\frac{1}{1+\frac{1}{4}}-0=\frac{4}{5} $ .
BETA
