Question

If A + B = 45°, prove that (1 + tan A) (1 + tan B) = 2 and hence deduce the value of tan 22`1/2`.

Answer 1

Given A + B = 45°

tan (A + B) = tan 45°

`(tan "A" + tan "B")/(1 - tan "A" tan "B")` = 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`1/2` in (1) we get

`(1 + tan 22 1/2) (1 + tan 22 1/2)` = 2

⇒ `(1 + tan 22 1/2)^2` = 2

⇒ 1 + tan 22`1/2 = +- sqrt2`

⇒ tan 22`1/2 = +- sqrt2 - 1`

Since 22`1/2` is acute, tan `22 1/2` is positive and therefore tan `22 1/2 = sqrt2 - 1`