If tan (A + B) = $$\sqrt{3}$$ and tan (A – B) = $$\dfrac{1}{\sqrt{3}}$$ ,0° < A + B ≤ 90°; A > B, find A and B.

Asked by Pragya Singh | 1 year ago |  87

##### Solution :-

tan (A + B) = $$\sqrt{3}$$

Since $$\sqrt{3}$$ = tan 60°

Now substitute the degree value

⇒ tan (A + B) = tan 60°

(A + B) = 60° … (i)

The above equation is assumed as equation (i)

tan (A – B) =$$\dfrac{1}{\sqrt{3}}$$

Since $$\dfrac{1}{\sqrt{3}}$$ = tan 30°

Now substitute the degree value

⇒ tan (A – B) = tan 30°

(A – B) = 30° … equation (ii)

Now add the equation (i) and (ii), we get

A + B + A – B = 60° + 30°

Cancel the terms B

2A = 90°

A= 45°

Now, substitute the value of A in equation (i) to find the value of B

45° + B = 60°

B = 60° – 45°

B = 15°

Therefore A = 45° and B = 15°

Answered by Abhisek | 1 year ago

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