# If `alpha+beta=pi/2 andbeta+phi=alpha ` then Prove that `Tan alpha = tan beta+2tan phi`

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llltkl | College Teacher | (Level 3) Valedictorian

Posted on

`alpha+beta=pi/2`

`rArr alpha=(pi/2-beta)` ------ (i)

and again, `alpha=(beta+phi)`  ---- (ii)

Therefore, from the relationship (ii), `tanalpha=tan(beta+phi)`

`rArrtanalpha=(tanbeta+tanphi)/(1-tanbeta*tanphi)`

`rArr tanalpha(1-tanbetatanphi)=tanbeta+tanphi`

`rArrtanalpha - tanalphatanbetatanphi=tanbeta+tanphi`

` `putting the value of alpha from relationship (i), we get

`tanalpha-tan(pi/2-beta)*tanbeta*tanphi=tanbeta+tanphi`

`rArrtanalpha-cotbeta*tanbeta*tanphi=tanbeta+tanphi`

`rArrtanalpha-(1/tanbeta)*tanbeta*tanphi=tanbeta+tanphi`

`rArrtanalpha-tanphi=tanbeta+tanphi`

`rArrtanalpha=tanbeta+2tanphi`

Hence the proof of the required relationship.

Sources:

aruv | High School Teacher | (Level 2) Valedictorian

Posted on

You have given

`alpha+beta=pi/2 and`

`beta+phi=alpha`

`tan(alpha)=tan(beta+phi)`

`=(tan(beta)+tan(phi))/(1-tan(beta)tan(phi))`

`tan(alpha){1-tan(beta)tan(phi)}=tan(beta)+tan(phi)`

`tan(alpha)-tan(alpha)tan(beta)tan(phi)=tan(beta)+tan(phi)`

`tan(alpha)-tan(pi/2-beta)tan(beta)tan(phi)=tan(beta)+tan(phi)`

`tan(alpha)-cot(beta)tan(beta)tan(phi)=tan(beta)+tan(phi)`

`tan(alpha)-tan(phi)=tan(beta)+tan(phi)`

`tan(alpha)=tan(beta)+tan(phi)+tan(phi)`

`tan(alpha)=tan(beta)+2tan(phi)`

``

Which is required result .

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