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

Prove that : $ \sin^{-1} \bigg(\large \frac{4}{5} \bigg)$$+\sin^{-1} \bigg(\large \frac{5}{13} \bigg) $$+ \sin^{-1} \bigg( \large\frac{16}{65} \bigg) = \large\frac{\pi}{2} $

Can you answer this question?
 
 

1 Answer

0 votes
Toolbox:
  • $\sin^{-1} x+\sin^{-1}y=\sin^{-1}\{x\sqrt{1-y^2}+y\sqrt{1-x^2}\}$
Step 1:
Given :$\sin^{-1}\bigg(\large\frac{4}{5}\bigg)$$+\sin^{-1}\bigg(\large\frac{5}{13}\bigg)$$+\sin^{-1}\bigg(\large\frac{16}{65}\bigg)$
By using the information from the toolbox we get,
$\Rightarrow \sin^{-1}\big[\large\frac{4}{5}$$\sqrt{1-(\large\frac{5}{13})^2}+\large\frac{5}{13}\sqrt{1-(\large\frac{4}{5})^2}\bigg]+\sin^{-1}\large\frac{16}{65}$
$\Rightarrow \sin^{-1}\bigg[\large\frac{4}{5}\times \frac{12}{13}+\frac{5}{13}\times \frac{3}{5}\bigg]+\sin^{-1}\big(\large\frac{16}{25}\big)$
$\Rightarrow \sin^{-1}\large\frac{63}{65}$$+\sin^{-1}\large\frac{16}{65}$
Step 2:
$\sin^{-1}\large\frac{63}{65}\big)$$=\cos^{-1}\sqrt{1-(\large\frac{63}{65})^2}=\cos^{-1}\large\frac{16}{65}$
$\Rightarrow \cos^{-1}\large\frac{16}{65}$$+\sin^{-1}\large\frac{16}{65}$
But $\sin^{-1}x+\cos^{-1}x=\large\frac{\pi}{2}$
$\Rightarrow \large\frac{\pi}{2}$
Hence proved.
answered Sep 23, 2013 by sreemathi.v
 

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
...