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Question

The value of $\frac{\tan 45^\circ - \tan 30^\circ}{1 + \tan 45^\circ \tan 30^\circ}$ is :

This question was previously asked in
RRB NTPC 2019 CBT 1 Question Paper PDF (31-Jan-2021) (Shift 1)
The correct answer is
$2 - \sqrt{3}$

Solving the Trigonometric Expression Value

The question asks for the value of the expression $\frac{\tan 45^\circ - \tan 30^\circ}{1 + \tan 45^\circ \tan 30^\circ}$.

We can solve this by substituting the known values of the tangent function for the standard angles $45^\circ$ and $30^\circ$.

  • $\tan 45^\circ = 1$
  • $\tan 30^\circ = \frac{1}{\sqrt{3}}$

Expression Evaluation Steps

Substitute these values into the given expression:

$ \frac{\tan 45^\circ - \tan 30^\circ}{1 + \tan 45^\circ \tan 30^\circ} = \frac{1 - \frac{1}{\sqrt{3}}}{1 + 1 \cdot \frac{1}{\sqrt{3}}} $

Simplify the numerator and the denominator:

$ \frac{\frac{\sqrt{3} - 1}{\sqrt{3}}}{\frac{\sqrt{3} + 1}{\sqrt{3}}} $

Cancel out the $\sqrt{3}$ term from the numerator and denominator:

$ \frac{\sqrt{3} - 1}{\sqrt{3} + 1} $

To simplify further, multiply the numerator and the denominator by the conjugate of the denominator, which is $(\sqrt{3} - 1)$:

$ \frac{(\sqrt{3} - 1)(\sqrt{3} - 1)}{(\sqrt{3} + 1)(\sqrt{3} - 1)} = \frac{(\sqrt{3} - 1)^2}{(\sqrt{3})^2 - 1^2} $

Expand the numerator and simplify the denominator:

$ \frac{(\sqrt{3})^2 - 2(\sqrt{3})(1) + 1^2}{3 - 1} = \frac{3 - 2\sqrt{3} + 1}{2} $

Combine the terms in the numerator:

$ \frac{4 - 2\sqrt{3}}{2} $

Divide both terms in the numerator by 2:

$ 2 - \sqrt{3} $

This matches option B.

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