Inverse Trigonometric Functions Question 107
Question: If $ {{\sin }^{-1}}1+{{\sin }^{-1}}\frac{4}{5}={{\sin }^{-1}}x, $ then what is x equal to?
Options:
A) 3/5
B) 4/5
C) 1
D) 0
Show Answer
Answer:
Correct Answer: A
Let $ {{\sin }^{-1}}(1)+si{n^{-1}}( \frac{4}{5} )={{\sin }^{-1}}x $
Let $ {{\sin }^{-1}}(1)=\theta \Rightarrow sin\theta =1\Rightarrow cos\theta =0 $
And $ {{\sin }^{-1}}( \frac{4}{5} )=\phi \Rightarrow \sin \phi =( \frac{4}{5} ) $
$ \Rightarrow \cos \phi =\sqrt{1-\frac{16}{25}}=\sqrt{\frac{9}{25}}=\frac{3}{5} $
$ \therefore {{\sin }^{-1}}x=\theta +\phi $
$ \Rightarrow x=\sin (\theta +\phi )=sin\theta cos\phi +cos\theta sin\phi $
$ =1\times \frac{3}{5}+0\times \frac{4}{5} $
$ \Rightarrow x=\frac{3}{5} $
BETA
