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JEE Main 2013

JEE Main 2013 Mathematics question on Trigonometry

The expression tan⁡A1−cot⁡A+cot⁡A1−tan⁡A\dfrac{\tan A}{1 - \cot A} + \dfrac{\cot A}{1 - \tan A} can be written as
  1. A.
    sin⁡Acos⁡A+1\sin A \cos A + 1
  2. B.
    sec⁡Acosec⁡A+1\sec A \operatorname{cosec} A + 1
  3. C.
    tan⁡A+cot⁡A\tan A + \cot A
  4. D.
    sec⁡A+cosec⁡A\sec A + \operatorname{cosec} A
Show answer and solution

Correct option: B

Solution

Let t = tan A, so cot A = 1/t.

First term: t/(1 - 1/t) = t/((t-1)/t) = t²/(t-1)

Second term: (1/t)/(1 - t) = (1/t)/(-(t-1)) = -1/(t(t-1))

Sum = t²/(t-1) - 1/(t(t-1)) = (t³ - 1)/(t(t-1)) = (t-1)(t²+t+1)/(t(t-1)) = (t²+t+1)/t

= t + 1 + 1/t = tan A + 1 + cot A

Now, tan A + cot A = sin A/cos A + cos A/sin A = 1/(sin A cos A) = sec A · cosec A.

So the expression equals sec A · cosec A + 1, which is option 2.

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