Question

if tan A = 1/ root 3 and tan b = root 3 prove that sin A cos B + cos A sin B = 1​

Answer 1

$lhs \: = sina.cosb + cosasinb \\ = \frac{sina.cosb + cosasinb}{cosa.cosb} \\ = tana + tanb \\ = \frac{1}{ \sqrt{3} } + \sqrt{3} \\ = \frac{1 + 3}{ \sqrt{3} } \\ = \frac{4}{ \sqrt{3} }$

Answer 2

$\tan \: a = \frac{1}{ \sqrt{3} }$

$\tan \: b = \sqrt{3}$

It is given that tan A = 1/√3

$\fcolorbox{red}{yellow}{tan30°\:=\:1/√3}$

on comparing we get,

$\bold\blue{A\:=\:30°}$

and tan B = √3

$\fcolorbox{red}{yellow}{tan60°\:=\:√3}$

$\bold\blue{B\:=\:60°}$

now,

= sin A cos B + cos A sin B

= sin 30° × cos 60° + cos 30° × sin 60°

= $\frac{1}{2} \times \frac{1}{2} \: + \: \frac{ \sqrt{3} }{2} \times \frac{ \sqrt{3} }{2}$

= $\frac{1}{4} + \frac{3}{4}$

= $\frac{1 + 3}{4}$

= $\frac{4}{4}$

= $1$

$\mathfrak\purple{please\:mark\:as\:brainliest.}$