1.

If tan A – tan B = x, cot B – cot A = y, Prove that cot (A – B) = \(\frac{1}{x}+\frac{1}{y}\)

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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