Question

if tan (a+b)=√3 and tan (a-b) =1/√3,a>b then value of a is​

Answer 1

Answer:

→ Value of a = 45°

★ GivEn :-

• $\sf \: \tan(a + b) = \sqrt{3}$
• $\sf \: \tan(a - b) = \dfrac{1}{\sqrt{3} }$
• a > b

★ To Find :-

• The value of a

Step-by-step explanation:

$\sf \: \tan(a + b) = \sqrt{3}$

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

Now, we know that,

$\sf \: \tan(60 \degree) = \sqrt{3}$

$\sf \: \tan(30 \degree) = \dfrac{1}{\sqrt{3} }$

Therefore,

$\sf \: a + b = 60 \\ \sf \: a - b = 30$

________________________

$\sf \: 2a = 90$

$\sf \implies \boxed{ \red{ a = 45}}$

Answer 2

Question

If tan(a+b)=√3 and tan (a-b) = 1/√3,a>b , then the value of a is ?

Solution

what is given ?

• it is given that tan(a+b)=√3
• and (a-b)=1/√3

what we have to find ?

• we have to find the value of a .

We know that tan(60°) is √3 and tan(30°) is 1/√3

Now,

a+b = 60 and a-b= 30

Therefore ,

2a=(a+b)+(a-b)

=> 2a=60°+30°

=> 2a=90°

=> a=90°/2

=> a=45 °

Hence , the value of a is 45° .