Inverse Trigonometric Functions Question 231
Question: If $ {{\sin }^{-1}}a+{{\sin }^{-1}}b+{{\sin }^{-1}}c=\pi , $ then the value of $ a\sqrt{(1-a^{2})}+b\sqrt{(1-b^{2})}+c\sqrt{(1-c^{2})} $ will be
[UPSEAT 1999]
Options:
A) $ 2abc $
B) $ abc $
C) $ \frac{1}{2}abc $
D) $ \frac{1}{3}abc $
Show Answer
Answer:
Correct Answer: A
Solution:
Let $ {{\sin }^{-1}}a=A, $
$ {{\sin }^{-1}}b=B, $
$ {{\sin }^{-1}}c=C $
$ \therefore \sin A=a,\sin B=b,\sin C=c $ and $ A+B+C=\pi , $ then $ \sin 2A+\sin 2B+\sin 2C $
$ =4\sin A\sin B\sin C $ -. .(i)
Therefore $ \sin A\cos A+\sin B\cos B+\sin C\cos C $ = $ 2\sin A\sin B\sin C $
Therefore $ \sin A\sqrt{(1-{{\sin }^{2}}A)}+\sin B\sqrt{(1-{{\sin }^{2}}B)}+\sin C\sqrt{1-{{\sin }^{2}}C} $
$ =2\sin A\sin B\sin C. $ ??(ii)
Therefore $ a\sqrt{(1-a^{2})}+b\sqrt{(1-b^{2})}+c\sqrt{{{(1-c)}^{2}}}=2abc $ , while $ {{\sin }^{-1}}a+{{\sin }^{-1}}b+{{\sin }^{-1}}c=\pi $ .
BETA
