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 = \(\boldsymbol{\rm x\over y}\)