Question

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

Answer 1

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) = \(\boldsymbol{\rm {1\over y} + {1\over x}}\)