Q103 / 180 · COMEDK 2022 · Mathematics

What is $\dfrac{\cos \theta}{1-\tan \theta}+\dfrac{\sin \theta}{1-\cot \theta}$ equal to?

  • A
    $\sin \theta-\cos \theta$
  • B
    $2 \sin \theta$
  • Correct answer
    C
    $\sin \theta+\cos \theta$
  • D
    $2 \cos \theta$

Solution

The given expression is $E = \dfrac{\cos \theta}{1 - \tan \theta} + \dfrac{\sin \theta}{1 - \cot \theta}$.

Substituting $\tan \theta = \dfrac{\sin \theta}{\cos \theta}$ and $\cot \theta = \dfrac{\cos \theta}{\sin \theta}$, we get:

$E = \dfrac{\cos \theta}{1 - \dfrac{\sin \theta}{\cos \theta}} + \dfrac{\sin \theta}{1 - \dfrac{\cos \theta}{\sin \theta}}$

$E = \dfrac{\cos \theta}{\dfrac{\cos \theta - \sin \theta}{\cos \theta}} + \dfrac{\sin \theta}{\dfrac{\sin \theta - \cos \theta}{\sin \theta}}$

$E = \dfrac{\cos^2 \theta}{\cos \theta - \sin \theta} + \dfrac{\sin^2 \theta}{\sin \theta - \cos \theta}$

Since $\sin \theta - \cos \theta = -(\cos \theta - \sin \theta)$, we have:

$E = \dfrac{\cos^2 \theta}{\cos \theta - \sin \theta} - \dfrac{\sin^2 \theta}{\cos \theta - \sin \theta}$

$E = \dfrac{\cos^2 \theta - \sin^2 \theta}{\cos \theta - \sin \theta}$

Using the identity $a^2 - b^2 = (a - b)(a + b)$, we get:

$E = \dfrac{(\cos \theta - \sin \theta)(\cos \theta + \sin \theta)}{\cos \theta - \sin \theta}$

$E = \cos \theta + \sin \theta$

Answer: $\sin \theta+\cos \theta$

Hence, the correct answer is option: 'C'.

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