Question

25) If tan(A + B) = v3 and tan(A - B) = 1/√3,0°< A+B S 90° find A and B.​

Answer 1

$\boxed{\bold{\large{GIVEN:}}}$

• $\to \sf tan(A+B) = \sqrt{3}$

• $\sf \to tan(A-B)=\dfrac{1}{\sqrt{3} }$

$\boxed{\bold{\large{TO \:\: FIND:}}}$

• $\sf A=?, B=?.$

$\boxed{\bold{\large{SOLUTION:}}}$

Simplify the given statements.

$\to \sf tan(A+B) = \sqrt{3}$

We know tanθ=√3 only when θ=60°

Here, tan(A+B) = √3. Thus, A+B=60°_____(1)

Similarly,

$\sf \to tan(A-B)=\dfrac{1}{\sqrt{3} }$

Here, tanθ=1/√3 only when θ=30°

Here, tan(A-B) = 1/√3. Thus, A-B=30°_____(1)

Add (1) and (2).

$\to \sf A+B = 60 ^{ o}\\\sf \to A-B=30^{o} \\----------\\2A=90^o$

$\boxed{\sf{\blue{A=45^o}}}$

Similarly, subtract (1) and (2).

$\sf A+B = 60 ^{ o}\\ \sf A-B=30^{o} \\ ----------\\2B=30^o$

$\boxed{\sf{\blue{B=15^o}}}$

∴A=45°, B=15°.