logo

Ask Questions, Get Answers

X
 

Prove that $\tan^{-1}{\frac{1}{4}}+\tan^{-1}{\frac{2}{9}}=\sin^{-1}{\frac{1}{\sqrt 5}}$

1 Answer

Toolbox:
  • $( tan^{-1}x+tan^{-1}y=tan^{-1}\frac{x+y}{1-xy}\:\:\:xy<1)$
  • $( tan^{-1}x=sin^{-1}\frac{x}{\sqrt{1+x^2}})$
L.H.S.=
By taking $(x=\frac{1}{4}\:and\:y=\frac{2}{9}\:we\:get)$
$(\frac{x+y}{1-xy}=\frac{\frac{1}{4}+\frac{2}{9}}{1-\frac{1}{4}.\frac{2}{9}}=\frac{17}{36}.\frac{36}{34}=\frac{17}{34}=\frac{1}{2})$
Substituting in the above formula we get L.H.S.=
$( tan^{-1}\frac{1}{4}+tan^{-1}\frac{2}{9}=tan^{-1}\frac{1}{2})$
By taking $(x=\frac{1}{2},\:\frac{x}{\sqrt{1+x^2}}=\frac{\frac{1}{2}}{\sqrt{1+\frac{1}{4}}}=\frac{1}{2}.\frac{2}{\sqrt{5}}=\frac{1}{\sqrt5})$
Substituting in the above formula of $(tan^{-1}x)$ we get
$(tan^{-1}\frac{1}{2}= sin^{-1}\frac{1}{\sqrt 5})$
=R.H.S.
answered Jul 9, 2014 by balaji.thirumalai
edited Nov 30 by meena.p
 
...