Question

If tan (a+b)= root 3 and tan(a-b)= 1/root3 , o degree < a+b _< 90 degree ; a>b , find a and b

Answer 1

Answer:

tan( a + b ) = √3

=> tan( a + b) = tan 60°

=> a + b = 60 ...... i

now,

tan( a - b) = 1/√3

=. > tan(a - b) = tan 30°

=> a - b = 30 ....... ii

on adding both equation

2a = 90

=> a = 45 °

putting a = 45 in (i)

a + b = 60

= > 45 + b = 60

=> b = 15°

Answer 2

$\huge\star{\green{\underline{\mathfrak{Answer :-}}}}$

$\leadsto \tan(a + b) = \sqrt{3}$ : EQUATION 1

$\leadsto \tan(a - b) = \dfrac{1}{ \sqrt{3} }$ : EQUATION 2

From equation 1,

$\leadsto\tan(a + b) = \tan(60)$

Because tan 60° is root 3.

So,

$\leadsto a + b = 60$ : EQUATION 3

From equation 2,

$\leadsto \tan(a - b) = \tan(30)$

Because tan 30° is 1 by root 3.

So,

$\leadsto a - b = 30$ : EQUATION 4

USING ELIMINATION METHOD TO SOLVE THE EQUATIONS 3 AND 4,

$\implies 2a = 90$

$\implies a = 45 \: \degree$

So, from Equation 3,

$\leadsto 45 + b = 60$

$\leadsto b = 15 \: \degree$

So, the angles are :-

$\huge{\boxed{\tt{a = \ {45}^{\circ} \ and \ b = {15}^{\circ}}}}$