Question

If A and B are acute angles such that tan ( A + B ) = √3 and
tan ( A – B ) = 1/(√3) then
pls answer this

Answer 1

Answer:

Given,

• tan ( A + B ) = √3
• tan ( A – B ) = 1/(√3)

$\small{And \: \: {0}^{0} \leqslant A \: and \: B \leqslant {90}^{0} }$

Now,

$tan ( A + B ) = √3$

$\implies tan ( A + B ) = tan \: {60}^{0}$

$\implies \: A + B = {60}^{0} - - - - - (1)$

And

$tan ( A – B ) = 1/(√3)$

$\implies \: tan ( A – B ) = tan \: {30}^{0}$

$\implies \: A – B \: = {30}^{0} - - - - - (2)$

$\large\bold{ \underline{Now, (1) + (2)}}$

$\implies \: A+B+A-B \: = {60}^{0} + {30}^{0}$

$\implies \: 2A = {90}^{0}$

$\implies \: \bold{ \boxed{ \color{blue}{A = {45}^{0} }}}$

$\large\bold{ \underline{ \mathtt{Putting \: \: the \: \: value \: of \: A \: \: in \: (1)}}}$

$\implies \: B \: = {60}^{0} - {45}^{0}$

$\implies \bold{ \boxed{ \mathtt{ \color{blue}{ \: B \: = {15}^{0} }}}}$