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