- #1

chwala

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- Homework Statement
- Prove that,

## 4\tan^{-1}\left[\dfrac{1}{5}\right]- \tan^{-1}\left[\dfrac{1}{239}\right]= \dfrac{π}{4}##

- Relevant Equations
- Trig. identities

I let,

## 4\tan^{-1}\left[\dfrac{1}{5}\right]- \tan^{-1}\left[\dfrac{1}{239}\right]= \dfrac{π}{4}##

##\tan^{-1}\left[\dfrac{1}{5}\right]- \dfrac{1}{4}\tan^{-1}\left[\dfrac{1}{239}\right]= \dfrac{π}{16}##

Then i let, ##\tan^{-1}\left[\dfrac{1}{5}\right] = α , \tan^{-1}\left[\dfrac{1}{239}\right]=β##

##⇒\tan α=\left[\dfrac{1}{5}\right],\tan β =\left[\dfrac{1}{239}\right], ##

##\tan (α-\dfrac{β}{4})= \left[\dfrac{\dfrac{1}{5}- \dfrac{1}{239×4}}{1+ \dfrac{1}{5}⋅\dfrac{1}{239×4}}\right]##

##\tan (α-\dfrac{β}{4})= \left[\dfrac{951}{4780} × \dfrac{4780}{4781}\right]##

##\tan (α-\dfrac{β}{4})=\left[\dfrac{951}{4781}\right]##

##\tan^{-1}(\tan (α-\dfrac{β}{4})≅11.25^0 = \dfrac{π}{16}##

##4[\tan^{-1}(\tan (α-\dfrac {β}{4})]≅45^0 = \dfrac{π}{4}##

I had a problem dealing with the ##4## in ##4\tan^{-1}\dfrac{1}{5}##... there may be a better approach...

## 4\tan^{-1}\left[\dfrac{1}{5}\right]- \tan^{-1}\left[\dfrac{1}{239}\right]= \dfrac{π}{4}##

##\tan^{-1}\left[\dfrac{1}{5}\right]- \dfrac{1}{4}\tan^{-1}\left[\dfrac{1}{239}\right]= \dfrac{π}{16}##

Then i let, ##\tan^{-1}\left[\dfrac{1}{5}\right] = α , \tan^{-1}\left[\dfrac{1}{239}\right]=β##

##⇒\tan α=\left[\dfrac{1}{5}\right],\tan β =\left[\dfrac{1}{239}\right], ##

##\tan (α-\dfrac{β}{4})= \left[\dfrac{\dfrac{1}{5}- \dfrac{1}{239×4}}{1+ \dfrac{1}{5}⋅\dfrac{1}{239×4}}\right]##

##\tan (α-\dfrac{β}{4})= \left[\dfrac{951}{4780} × \dfrac{4780}{4781}\right]##

##\tan (α-\dfrac{β}{4})=\left[\dfrac{951}{4781}\right]##

##\tan^{-1}(\tan (α-\dfrac{β}{4})≅11.25^0 = \dfrac{π}{16}##

##4[\tan^{-1}(\tan (α-\dfrac {β}{4})]≅45^0 = \dfrac{π}{4}##

I had a problem dealing with the ##4## in ##4\tan^{-1}\dfrac{1}{5}##... there may be a better approach...

Last edited: