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If cos A = 0.5 and cos B = 12\dfrac{1}{\sqrt2} ; find the value of : tan A−tan B1+tan A tan B\dfrac{\text{tan A} - \text{tan B}}{1 + \text{tan A } \text{tan B}}.

Trigonometrical Ratios

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Answer

cos A = 0.5

cos A = 510\dfrac{5}{10} = 12\dfrac{1}{2}

⇒ cos A = cos 60°

⇒ A = 60°

cos B = 12\dfrac{1}{\sqrt2}

⇒ cos B = cos 45°

⇒ B = 45°

Now, find the value of

=tan A−tan B1+tan A tan B=tan 60°−tan 45°1+tan 60°tan 45°=3−11+3×1=3−11+3=3−11+3×1−31−3=(3−1)×(1−3)(1+3)×(1−3)=3×(1−3)−1×(1−3)1−(3)2=3−3−1+31−3=23−4−2=2−3= \dfrac{\text{tan A} - \text{tan B}}{1 + \text{tan A } \text{tan B}}\\[1em] = \dfrac{\text{tan 60°} - \text{tan 45°}}{1 + \text{tan 60°} \text{tan 45°}}\\[1em] = \dfrac{\sqrt{3} - 1}{1 + \sqrt{3} \times 1}\\[1em] = \dfrac{\sqrt{3} - 1}{1 + \sqrt{3}}\\[1em] = \dfrac{\sqrt{3} - 1}{1 + \sqrt{3}} \times \dfrac{1 - \sqrt{3}}{1 - \sqrt{3}}\\[1em] = \dfrac{(\sqrt{3} - 1) \times (1 - \sqrt{3})}{(1 + \sqrt{3}) \times (1 - \sqrt{3})}\\[1em] = \dfrac{\sqrt{3} \times (1 - \sqrt{3}) - 1 \times (1 - \sqrt{3})}{1 - (\sqrt{3})^2 }\\[1em] = \dfrac{\sqrt{3} - 3 - 1 + \sqrt{3}}{1 - 3}\\[1em] = \dfrac{2\sqrt{3} - 4}{- 2}\\[1em] = 2 - \sqrt{3}

Hence, the value of tan A−tan B1+tan A tan B=2−3\dfrac{\text{tan A} - \text{tan B}}{1 + \text{tan A } \text{tan B}} = 2 - \sqrt{3}.

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