Given A = 60° and B = 30°, prove that : tan (A - B) = tan A - tan B/1 + tan A.tan B

L.H.S. = tan (A - B) = tan (60° - 30°) = tan 30° = ∴ L.H.S. = R.H.S. Hence, .

Mathematics

Given A = 60° and B = 30°, prove that :
$\text{tan (A - B)} = \dfrac{\text{tan A - tan B}}{\text{1 + \text{ tan A }} . \text{ tan B}}$

Trigonometric Identities

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Answer

$\text{tan (A - B)} = \dfrac{\text{tan A - tan B}}{\text{1 + tan} \text{ A } . \text{ tan B}}$

L.H.S. = tan (A - B) = tan (60° - 30°)

= tan 30° = $\dfrac{1}{\sqrt3}$

$\text{R.H.S.} = \dfrac{\text{tan A - tan B}}{\text{1 + tan} \text{ A } . \text{ tan B}}\\[1em] = \dfrac{\text{tan 60° - tan 30°}}{\text{1 + tan} \text{ 60° } . \text{ tan 30°}}\\[1em] = \dfrac{\sqrt3 - \dfrac{1}{\sqrt3}}{1 + \sqrt3 \times \dfrac{1}{\sqrt3}}\\[1em] = \dfrac{\dfrac{\sqrt3 \times \sqrt3}{\sqrt3} - \dfrac{1}{\sqrt3}}{1 + \cancel{\sqrt3} \times \dfrac{1}{\cancel{\sqrt3}}}\\[1em] = \dfrac{\dfrac{3}{\sqrt3} - \dfrac{1}{\sqrt3}}{1 + 1}\\[1em] = \dfrac{\dfrac{3 - 1}{\sqrt3}}{1 + 1}\\[1em] = \dfrac{\dfrac{2}{\sqrt3}}{2}\\[1em] = \dfrac{\dfrac{\cancel{2}}{\sqrt3}}{\cancel{2}}\\[1em] = \dfrac{1}{\sqrt3}$

∴ L.H.S. = R.H.S.

Hence, $\text{tan (A - B)} = \dfrac{\text{tan A - tan B}}{\text{1 + tan} \text{ A } . \text{ tan B}}$ .

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Mathematics

Given A = 60° and B = 30°, prove that :
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Trigonometric Identities

Answer

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