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JEE Main 20137 April
Exam
JEE Main
Year
2013
Date
7 April
Subject
Mathematics
Chapter
Trigonometric Ratios & Identities

JEE Main 2013 · 7 April · Mathematics · Trigonometric Ratios & Identities

The expression (tanA)/(1 - cotA) + (cotA)/(1 - tanA) can be written...

The expression tanA1−cotA+cotA1−tanA\dfrac{tanA}{1 - cotA} + \dfrac{cotA}{1 - tanA} can be written as :

Options

  1. AtanA+cotAtanA + cotA
  2. BsecA+cosecAsecA + cosecA
  3. CsinAcosA+1sinAcosA + 1
  4. DsecAcosecA+1secAcosecA + 1

Correct answer

D. secAcosecA+1secAcosecA + 1

Step-by-step solution

Given tanA1−cotA+cotA1−tanA\dfrac{tanA}{1 - cotA} + \dfrac{cotA}{1 - tanA}

=sin⁡2AcosA(sinA−cosA)+cos⁡2AsinA(cosA−sinA)= \dfrac{\sin^{2}A}{cosA\left(sinA - cosA\right)} + \dfrac{\cos^{2}A}{sin A \left(cosA - sinA\right)}

=sin⁡3A−cos⁡3AsinAcosA(sinA−cosA)= \dfrac{\sin^{3}A - \cos^{3}A}{sinAcosA\left(sinA - cosA\right)}

=sin⁡2A+cos⁡2A+sinAcosAsinAcosA= \dfrac{\sin^{2}A + \cos^{2}A + sinAcosA}{sinAcosA}

=secAcosecA+1= secAcosecA + 1
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