If tan A - tan B = x and cot B - cot A = y, then cot (A - B) is equal

1.	If tan A - tan B = x and cot B - cot A = y, then cot (A - B) is equal to1. \(\dfrac{1}{x}+\dfrac{1}{y}\)2. \(\dfrac{1}{x}-\dfrac{1}{y}\)3. \(-\dfrac{1}{x}+\dfrac{1}{y}\)4. \(-\dfrac{1}{x}-\dfrac{1}{y}\)
Answer» Correct Answer - Option 1 : \(\dfrac{1}{x}+\dfrac{1}{y}\) Concept: The identities of trigonometry are: \(\rm \tan a={\sin a\over \cos a}\) \(\rm \cot a ={\cos a\over\sin a} = {1\over \tan a}\) \(\rm \tan (a+b)={\tan a +\tan b\over{1-\tan a \tan b}}\) \(\rm \tan (a-b)={\tan a -\tan b\over{1+\tan a \tan b}}\) Calculation: Given cot B - cot A = y ⇒ \(\rm {1\over\tan B}-{1\over\tan A} =y\) ⇒ \(\rm {\tan A - \tan B\over\tan A\tan B} =y\) Given tan A - tan B = x ⇒ \(\rm {x\over\tan A\tan B} =y\) ⇒ \(\boldsymbol{\rm \tan A\tan B ={x\over y}}\) Now, \(\rm \tan (A-B)={\tan A -\tan B\over{1+\tan A \tan B}}\) ⇒ tan (A - B) = \(\rm {x\over{1+{x\over y}}}\) ⇒ tan (A - B) = \(\rm {xy\over{x+y}}\) cot (A - B) = \(\rm 1\over\tan (A - B)\) ⇒ cot (A - B) = \(\rm x+y\over xy\) ⇒ cot (A - B) = \(\boldsymbol{\rm {1\over y} + {1\over x}}\)

If tan A - tan B = x and cot B - cot A = y, then cot (A - B) is equal to1. \(\dfrac{1}{x}+\dfrac{1}{y}\)2. \(\dfrac{1}{x}-\dfrac{1}{y}\)3. \(-\dfrac{1}{x}+\dfrac{1}{y}\)4. \(-\dfrac{1}{x}-\dfrac{1}{y}\)

Answer» Correct Answer - Option 1 : \(\dfrac{1}{x}+\dfrac{1}{y}\)

Concept:

The identities of trigonometry are:

Calculation:

Given

cot B - cot A = y

⇒ \(\rm {1\over\tan B}-{1\over\tan A} =y\)

⇒ \(\rm {\tan A - \tan B\over\tan A\tan B} =y\)

Given tan A - tan B = x

⇒ \(\rm {x\over\tan A\tan B} =y\)

⇒ \(\boldsymbol{\rm \tan A\tan B ={x\over y}}\)

Now, \(\rm \tan (A-B)={\tan A -\tan B\over{1+\tan A \tan B}}\)

⇒ tan (A - B) = \(\rm {x\over{1+{x\over y}}}\)

⇒ tan (A - B) = \(\rm {xy\over{x+y}}\)

cot (A - B) = \(\rm 1\over\tan (A - B)\)

⇒ cot (A - B) = \(\rm x+y\over xy\)

⇒ cot (A - B) = \(\boldsymbol{\rm {1\over y} + {1\over x}}\)