# Basic math

if tan alpha =1/3 and tan beta =1/7 prove that tan(2 alpha+beta)=45 degrees

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1. tan ( A + B ) = ( tan A + tan B ) / ( 1 - tan A * tan B )

In this case:

A = 2 alpha , B = beta

so

tan ( 2 alpha + beta ) = [ tan ( 2 alpha ) + tan ( beta ) ] / [ 1 - tan ( 2 alpha ) * tan ( beta ) ]

Since the:

tan ( 2 alpha ) = 2 * tan ( alha ) / [ 1 - tan ^ 2 ( alpha ) ]

put:

tan ( alpha ) = 1 / 3 in this formula

tan ( 2 alpha ) = 2 * ( 1 / 3) / [ 1 - ( 1 / 3) ^ 2 ]

tan ( 2 alpha ) = ( 2 / 3 ) / ( 1 - 1 / 9 )

tan ( 2 alpha ) = ( 2 / 3 ) / ( 9 / 9 - 1 / 9 )

tan ( 2 alpha ) = ( 2 / 3 ) / ( 8 / 9 )

tan ( 2 alpha ) = ( 2 * 9 ) / ( 8 * 3 )

tan ( 2 alpha ) = 18 / 24

tan ( 2 alpha ) = 6 * 3 / ( 6 * 4 )

tan ( 2 alpha ) = 3 / 4

Replace:

tan ( 2 alpha ) = 3 / 4 and tan ( beta ) = 1 / 7 in formula

tan ( 2 alpha + beta ) = [ tan ( 2 alpha ) + tan ( beta ) ] / [ 1 - tan ( 2 alpha ) * tan ( beta ) ]

tan ( 2 alpha + beta ) = ( 3 / 4 + 1 / 7 ) / [ 1 - ( 3 / 4 ) * ( 1 / 7 ) ]

tan ( 2 alpha + beta ) = [ 3 * 7 / ( 4 * 7 ) + 1 * 4 / ( 7 * 4 ) ] / ( 1 - 3 / 28 )

tan ( 2 alpha + beta ) = ( 21 / 28 + 4 / 28 ) / [ 28 / 28 - 3 / 28 ) ]

tan ( 2 alpha + beta ) = ( 25 / 28 ) / (25 / 28 )

tan ( 2 alpha + beta ) = 1

tan ( 2 alpha + beta ) = tan ( 45 ° )

becouse:

tan ( 45 ° ) = 1

So:

tan ( 2 alpha + beta ) = tan ( 45 ° )

2 alpha + beta = 45 °

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