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Payal Eddie Ramesh
Asked: 2022-11-11T00:01:13+05:30 In:

If \(\rm \cot\left[\tan^{-1}{a} – \tan^{-1}{\left({x-y\over x+y}\right)}\right] = 1\), then
1. a = \(\rm y\over x\)
2. a = \(\rm x\over y\)
3. a = \(\rm x-y\over x\)
4. a = \(\rm x+y\over x-y\)

If \(\rm \cot\left[\tan^{-1}{a} – \tan^{-1}{\left({x-y\over x+y}\right)}\right] = 1\), then
1. a = \(\rm y\over x\)
2. a = \(\rm x\over y\)
3. a = \(\rm x-y\over x\)
4. a = \(\rm x+y\over x-y\)
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  1. Correct Answer – Option 2 : a = \(\rm x\over y\)

    Concept:

    Inverse trigonometric identity

    • \(\rm \tan^{-1}{a} + \tan^{-1}b = \tan^{-1}{\left(a+b\over1-ab\right)}\)
    • \(\rm \tan^{-1}{a} – \tan^{-1}b = \tan^{-1}{\left(a-b\over1+ab\right)}\)
    • sin-1(sin a) = a
    • cos-1(cos a) = a
    • tan-1(tan a) = a
    • cot-1(cot a) = a

    Calculation:

    S = \(\rm \cot\left[\tan^{-1}{a} – \tan^{-1}{\left({x-y\over x+y}\right)}\right] = 1\)

    ⇒ \(\rm \cot\left[\tan^{-1}{a} – \tan^{-1}{\left({x-y\over x+y}\right)}\right] = \cot45\) (∵ cot 45 =1)

    Taking cot-1 on both sides

    ⇒ \(\rm \cot^{-1}\cot\left[\tan^{-1}{a} – \tan^{-1}{\left({x-y\over x+y}\right)}\right] = \cot^{-1}(\cot45)\)

    ⇒ \(\rm \tan^{-1}{a} – \tan^{-1}{\left({x-y\over x+y}\right)} = 45\)

    ⇒ \(\rm \tan^{-1}\left[{a – {\left({x-y\over x+y}\right)}\over1+a\times{\left({x-y\over x+y}\right)}}\right] = 45\)

    Taking tan on both sides

    ⇒ \(\rm \tan \tan^{-1}\left[{a – {\left({x-y\over x+y}\right)}\over1+a{\left({x-y\over x+y}\right)}}\right] = \tan45\)

    ⇒ \(\rm {a – {\left({x-y\over x+y}\right)}\over1+a{\left({x-y\over x+y}\right)}} = 1\)

    ⇒ \(\rm {a – {\left({x-y\over x+y}\right)}=1+a{\left({x-y\over x+y}\right)}}\)

    ⇒ \(\rm {a {\left(1-{x-y\over x+y}\right)}=1+{\left({x-y\over x+y}\right)}}\)

    ⇒ a(x + y – (x – y)) = x + y + x – y

    ⇒ 2ya = 2x

    ⇒ a = \(\boldsymbol{\rm x\over y}\)

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