Value of \[\tan (\frac{ 1 }{ 2 } \cos ^{-1}\frac{ \sqrt{5} }{3 })\]
@Yahoo! first find the value of \[\tan ( \cos ^{-1}\frac{ \sqrt{5} }{3 })\] could you find it?
\[\tan ( \tan ^{-1} \frac{ 2 }{ \sqrt{5} })\]
@Yahoo! I can see that you have closed your previous question before it was done. Please don't do this. It's against the code of conduct http://openstudy.com/code-of-conduct Thanks
lol...@ash2326 i got...that answer....and was waiting...for...Sauravs method..
@sauravshakya Any Idea
@ash2326 ...()
CANT WE USE CALCULATOR?
No Calculator.)
let \[x=\cos^{-1} (\frac{\sqrt 5}{3})\] so we got \[\tan x=\frac{2}{\sqrt 5}\] now find \[\tan (\frac x 2)\] we know \[\large \tan x=\frac{2\tan \frac x 2}{1-\tan^2 \frac x2}\] now find tan x/2 from this
i think....this may be little easier: \[\cos ^{-1} \frac{ \sqrt{5} }{ 3 } = 2x\]
it doesn't matter, you have to solve a quadratic at last. Substitute tan x/2 or tan x= y
i got this...)....
Which one shuld i choose
Both of them r in my option
the one which is positive
Why...
\[\frac{ \pm3 - \sqrt{5} }{ 2 }\]
since \[x=\cos^{-1}\frac{\sqrt{5}}{3} \] \[\text{x will lie in the range 0 to}\frac{\pi}{2} \text{since }\frac{\sqrt{5}}{3} \text{is positive }\] so \[\text{x/2 will also lie in the range 0 to}\frac{\pi}{2}\] \[\large \text{tangent is positive in this range :)}\]
Got.....it............thxxxxx
welcome :)
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