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Pravin Rajesh Mander
Asked: 2022-11-11T00:47:14+05:30 In:

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. \(-\dfrac{1}{x}-\dfrac{1}{y}\)

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

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