`alphaa n dbeta` are the positive acute angles and satisfying equation `5sin2beta=3s in2alphaa n dtanbeta=3tanalpha` simultaneously. Then the value of `tanalpha+tanbeta` is _________

Answer

4

Explanation

`5 xx (2 tanbeta)/( 1+ tan^(2) beta ) = 3xx ( 2tan alpha)/(1+tan^(2) alpha )`
or ` ( 5tan beta)/( 1+ tan^(2) beta ) = (3 tan alpha )/( 1+ tan^(2) alpha)`
Substituting `tan beta = 3 tan alpha`, we have
`" " ( 5xx 3 tan alpha ) /( 1+9 tan^(2) alpha ) = ( 3tan alpha )/( 1+ tan^(2) alpha)`
or `5 +5 tan^(2) alpha = 1+ 9 tan ^(2) alpha `
or `4 tan ^(2) alpha = 4`
or `tan alpha = 1`, i.e., `tan beta = 3`
`therefore tan alpha + tan beta = 4`