logo

Ask Questions, Get Answers

 
X
 Search
Want to ask us a question? Click here
Browse Questions
Ad
Home  >>  CBSE XII  >>  Math  >>  Model Papers
0 votes

Show that $ \sin^{-1}\large\frac{12}{13}$$+\cos^{-1}\large\frac{4}{5}$$+\tan^{-1}\large\frac{63}{16} $$= \pi $

Can you answer this question?
 
 

1 Answer

0 votes
Toolbox:
  • \( sin^{-1}\large\frac{12}{13}=tan^{-1}\large\frac{12}{5}\)
  • \( cos^{-1}\large\frac{4}{5}=tan^{-1}\large\frac{3}{4}\)
  • \( tan^[-1}x+tan^{-1}y=tan^{-1} \bigg( \large\frac{x+y]{1-xy} \bigg)
  • \( xy > 1\)
\( sin^{-1}\large\frac{12}{13}+cos^{-1}\large\frac{4}{5}+tan^{-1}\large\frac{63}{16}\)
\( = tan^{-1}\large\frac{12}{5}+tan^{-1}\large\frac{3}{4}+tan6{-1}\large\frac{63}{13}\)
\( = \pi+tan^{-1} \bigg[ \large\frac{\large\frac{12}{5}+\large\frac{3}{4}}{1-\large\frac{36}{20}} \bigg] +tan^{-1}\large\frac{63}{13}\)
\( = \pi+tan^{-1}\large\frac{63}{-13}+tan^{-1}\large\frac{63}{13}\)
\( = \pi = R.H.S\)

 

answered Feb 28, 2013 by thanvigandhi_1
edited Mar 19, 2013 by thanvigandhi_1
 

Related questions

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
...