logo

Ask Questions, Get Answers

 
X
 Search
Want to ask us a question? Click here
Browse Questions
Ad
0 votes

The value of $\sin(\tan^{-1}(.75))$ is equal to

\[(A)\quad.75\quad(B)\quad1.5\quad(C)\quad.96\quad(D)\quad\sin1.5\]
Can you answer this question?
 
 

1 Answer

0 votes
Toolbox:
  • \(2tan^{-1}x=tan^{-1}\large\frac{2x}{1-x^2},\:\:|x|<1\)
  • \(tan^{-1}x=sin^{-1}\large\frac{x}{\sqrt{1+x^2}}\)
Ans (C) .96
By taking \(x=.75=\large\frac{3}{4}\),
\(\large\frac{2x}{1-x^2}=\large\frac{2.\large\frac{3}{4}}{1-\large\frac{9}{16}}=\large\frac{3}{2}.\large\frac{16}{7}=\large\frac{24}{7}\)
 
Substituting in the given expression
\( sin \bigg[ 2tan^{-1}(.75) \bigg] = sin \bigg( tan^{-1}\large\frac{24}{7} \bigg)\)
 
By taking \(x=\large\frac{24}{7}\), we get
\(\large\frac{x}{\sqrt{1+x^2}}=\large\frac{\large\frac{24}{7}}{\sqrt{1+\large\frac{576}{49}}}=\large\frac{24}{7}.\large\frac{7}{25}=\large\frac{24}{25}\)
 
\(\Rightarrow\:tan^{-1}\large\frac{24}{7}=sin^{-1}\big( \large\frac{24}{25} \big)\)
\(\Rightarrow\: sin \bigg( tan^{-1}\large\frac{24}{7}\bigg)= sin\bigg(sin^{-1}\large \frac{24}{25} \bigg) =\large\frac{24}{25}=.9

 

answered Feb 18, 2013 by thanvigandhi_1
edited Mar 16, 2013 by thanvigandhi_1
 
Ask Question
student study plans
x
JEE MAIN, CBSE, NEET Mobile and Tablet App
The ultimate mobile app to help you crack your examinations
Get the Android App
...