is equal to
A)
B)
C)
D)
Correct Answer: D
Solution :
Let \[{{\cot }^{-1}}\,(\sqrt{2}-1)=\theta \Rightarrow \,\,cot\,\theta \,\,=\,\,\sqrt{2}-1\] \[\therefore \,\,{{\cos }^{-1}}(cos\,2\theta )=co{{s}^{-1}}\left( \frac{1-{{\tan }^{2}}\theta }{1+{{\tan }^{2}}\theta } \right)\] \[\therefore \,\,{{\cos }^{-1}}\left( \frac{{{\cot }^{2}}\theta -1}{{{\cot }^{2}}\theta +1} \right)=co{{s}^{-1}}\,\left( \frac{2-2\sqrt{2}}{4-2\sqrt{2}} \right)\] \[=\,\,{{\cos }^{-1}}\left( -\frac{1}{\sqrt{2}} \right)=\frac{3\pi }{4}\]
You need to login to perform this action.
You will be redirected in
3 sec
