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If tan A – tan B = x, cot B – cot A = y, Prove that cot (A – B) = \(\frac{1}{x}+\frac{1}{y}\) |
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Answer» Given, tan A – tan B = x ...(i) and cot B – cot A = y ...(ii) From (ii), cot B – cot A = y ⇒ \(\frac{1}{tanB}-\frac{1}{tanA}=y\) ⇒ \(\frac{tanA - tanB}{tanAtanB}=y\) ⇒ \(\frac{x}{tanAtanB}=y\) ⇒ tanA tanB = \(\frac{x}{y}\) Now, ⇒ cot (A − B) = \(\frac{1}{tan(A-B)}\)= \(\frac{1+tanAtanB}{tanA-tanB}\) ⇒ cot (A − B) = \(\frac{1+\frac{x}{y}}{x}\)= \(\frac{y+x}{xy}\) ⇒ cot (A − B) = \(\frac{1}{x}+\frac{1}{y}\)Hence Proved |
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