If A + B = 45°, prove that (1 + tan A) (1 + tan B) = 2 and hence dedu

1.	If A + B = 45°, prove that (1 + tan A) (1 + tan B) = 2 and hence deduce the value of tan 22\(\frac{1}{2}\).
Answer» Given A + B = 45° tan (A + B) = tan 45° \(\frac{tanA+tanB}{1-tanAtanB}\) = 1 tan A + tan B = 1 – tan A.tan B tan A + tan B + tan A tan B = 1 Add 1 on both sides we get, (1 + tan A) + tan B + tan A tan B = 2 1(1+ tan A) + tan B (1 + tan A) = 2 (1 + tan A) (1 + tan B) = 2 … (1) Put A = B = 22\(\frac{1}{2}\) in (1) we get (1 + tan 22\(\frac{1}{2}\)) (1 + tan 22\(\frac{1}{2}\)) = 2 ⇒ (1 + tan 22\(\frac{1}{2}\))² = 2 ⇒ 1 + tan 22\(\frac{1}{2}\) = ±√2 ⇒ tan 22\(\frac{1}{2}\) = ±√2 – 1 Since 22\(\frac{1}{2}\) is acute, tan 22\(\frac{1}{2}\) is positive and Therefore tan 22\(\frac{1}{2}\) = √2 – 1

If A + B = 45°, prove that (1 + tan A) (1 + tan B) = 2 and hence deduce the value of tan 22\(\frac{1}{2}\).

Answer»

Given A + B = 45°

tan (A + B) = tan 45°

\(\frac{tanA+tanB}{1-tanAtanB}\) = 1

tan A + tan B = 1 – tan A.tan B

tan A + tan B + tan A tan B = 1

Add 1 on both sides we get,

(1 + tan A) + tan B + tan A tan B = 2

1(1+ tan A) + tan B (1 + tan A) = 2

(1 + tan A) (1 + tan B) = 2 … (1)

Put A = B = 22\(\frac{1}{2}\) in (1) we get

(1 + tan 22\(\frac{1}{2}\)) (1 + tan 22\(\frac{1}{2}\)) = 2

⇒ (1 + tan 22\(\frac{1}{2}\))² = 2

⇒ 1 + tan 22\(\frac{1}{2}\) = ±√2

⇒ tan 22\(\frac{1}{2}\) = ±√2 – 1

Since 22\(\frac{1}{2}\) is acute, tan 22\(\frac{1}{2}\) is positive and

Therefore tan 22\(\frac{1}{2}\) = √2 – 1