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Question

The value of \({\rm{tan}}\left( {2{{\tan }^{ - 1}}\frac{1}{5} - \frac{\pi }{4}} \right)\) is

The correct answer is \(- \frac{7}{{17}}\)

Understanding the Trigonometric Expression

We need to find the value of the given trigonometric expression: \({\rm{tan}}\left( {2{{\tan }^{ - 1}}\frac{1}{5} - \frac{\pi }{4}} \right)\). This expression involves the tangent function applied to a difference of two angles, one of which is related to an inverse tangent function and the other is a standard angle.

Let the expression be denoted by \(E\). We can write it in the form \(E = \tan(A - B)\), where \(A = 2{{\tan }^{ - 1}}\frac{1}{5}\) and \(B = \frac{\pi }{4}\).

Applying Relevant Trigonometric Identities

To evaluate this, we will use two important trigonometric identities:

  1. The identity for the tangent of the difference of two angles: \[{\rm{tan}}(X - Y) = \frac{{{\rm{tan}}X - {\rm{tan}}Y}}{{1 + {\rm{tan}}X{\rm{tan}}Y}}\]
  2. The identity for twice the inverse tangent: \[2{{\tan }^{ - 1}}x = {{\tan }^{ - 1}}\frac{{2x}}{{1 - {x^2}}}, \text{ for } |x| < 1\]

Step-by-Step Calculation

Step 1: Simplify the term involving \(2{{\tan }^{ - 1}}\)

Let's first simplify \(A = 2{{\tan }^{ - 1}}\frac{1}{5}\). Here, \(x = \frac{1}{5}\). Since \(|\frac{1}{5}| = \frac{1}{5} < 1\), we can use the identity \(2{{\tan }^{ - 1}}x = {{\tan }^{ - 1}}\frac{{2x}}{{1 - {x^2}}}\).

Substituting \(x = \frac{1}{5}\):

\[2{{\tan }^{ - 1}}\frac{1}{5} = {{\tan }^{ - 1}}\frac{{2 \times \frac{1}{5}}}{{1 - {{\left(\frac{1}{5}\right)}^2}}}\] \[ = {{\tan }^{ - 1}}\frac{{\frac{2}{5}}}{{1 - \frac{1}{{25}}}}\] \[ = {{\tan }^{ - 1}}\frac{{\frac{2}{5}}}{{\frac{{25 - 1}}{{25}}}}\] \[ = {{\tan }^{ - 1}}\frac{{\frac{2}{5}}}{{\frac{{24}}{{25}}}}\] \[ = {{\tan }^{ - 1}}\left(\frac{2}{5} \times \frac{25}{24}\right)\] \[ = {{\tan }^{ - 1}}\left(\frac{2 \times 25}{5 \times 24}\right)\] \[ = {{\tan }^{ - 1}}\left(\frac{50}{120}\right)\] \[ = {{\tan }^{ - 1}}\frac{5}{12}\]

So, \(A = {{\tan }^{ - 1}}\frac{5}{12}\).

Step 2: Express \(\frac{\pi}{4}\) in terms of inverse tangent

We know that \(\tan\left(\frac{\pi}{4}\right) = 1\). Therefore, \(B = \frac{\pi}{4}\) can be written as \({{\tan }^{ - 1}}1\).

So, the original expression becomes \(E = {\rm{tan}}\left( {{{\tan }^{ - 1}}\frac{5}{12} - {{\tan }^{ - 1}}1} \right)\).

Step 3: Apply the tangent difference identity

Now we have the expression in the form \({\rm{tan}}(A - B)\), where \(A = {{\tan }^{ - 1}}\frac{5}{12}\) and \(B = {{\tan }^{ - 1}}1\). Using the identity \({\rm{tan}}(X - Y) = \frac{{{\rm{tan}}X - {\rm{tan}}Y}}{{1 + {\rm{tan}}X{\rm{tan}}Y}}\):

Let \(X = {{\tan }^{ - 1}}\frac{5}{12}\) and \(Y = {{\tan }^{ - 1}}1\). Then \(\tan X = \frac{5}{12}\) and \(\tan Y = 1\).

\[E = \frac{{\tan \left({{\tan }^{ - 1}}\frac{5}{12}\right) - \tan \left({{\tan }^{ - 1}}1\right)}}{{1 + \tan \left({{\tan }^{ - 1}}\frac{5}{12}\right) \times \tan \left({{\tan }^{ - 1}}1\right)}}\] \[E = \frac{{\frac{5}{12} - 1}}{{1 + \frac{5}{12} \times 1}}\] \[E = \frac{{\frac{5 - 12}{12}}}{{1 + \frac{5}{12}}}\] \[E = \frac{{\frac{-7}{12}}}{{\frac{12 + 5}{12}}}\] \[E = \frac{{\frac{-7}{12}}}{{\frac{17}{12}}}\] \[E = \frac{-7}{12} \times \frac{12}{17}\] \[E = \frac{-7}{17}\]

The value of the expression \({\rm{tan}}\left( {2{{\tan }^{ - 1}}\frac{1}{5} - \frac{\pi }{4}} \right)\) is \(-\frac{7}{17}\).

Summary of Steps

Here is a quick summary of the steps taken:

  1. Identified the structure of the expression as \(\tan(A-B)\).
  2. Used the identity \(2{{\tan }^{ - 1}}x = {{\tan }^{ - 1}}\frac{{2x}}{{1 - {x^2}}}\) to simplify \(2{{\tan }^{ - 1}}\frac{1}{5}\) to \({{\tan }^{ - 1}}\frac{5}{12}\).
  3. Expressed \(\frac{\pi}{4}\) as \({{\tan }^{ - 1}}1\).
  4. Applied the tangent difference formula \({\rm{tan}}(X - Y) = \frac{{{\rm{tan}}X - {\rm{tan}}Y}}{{1 + {\rm{tan}}X{\rm{tan}}Y}}\).
  5. Substituted the values and performed algebraic simplification to arrive at the final result.

Revision Table: Trigonometric Identities

Identity Type Formula Notes
Tangent Difference \({\rm{tan}}(X - Y) = \frac{{{\rm{tan}}X - {\rm{tan}}Y}}{{1 + {\rm{tan}}X{\rm{tan}}Y}}\) Used for simplifying \(\tan(A-B)\) form.
Inverse Tangent (Double Angle) \(2{{\tan }^{ - 1}}x = {{\tan }^{ - 1}}\frac{{2x}}{{1 - {x^2}}}\) Valid for \(|x| < 1\).
Inverse Tangent Property \(\tan({{\tan }^{ - 1}}x) = x\) Used when applying tangent function to its inverse.

Additional Information: Domain and Range of Inverse Tangent

The inverse tangent function, \({{\tan }^{ - 1}}x\) or \(\arctan x\), has a specific domain and range that are important to remember when working with inverse trigonometric functions:

  • Domain: The domain of \({{\tan }^{ - 1}}x\) is all real numbers, \((-\infty, \infty)\).
  • Range: The range of \({{\tan }^{ - 1}}x\) is \(\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)\). This means the output of \({{\tan }^{ - 1}}x\) is an angle strictly between \(-\frac{\pi}{2}\) and \(\frac{\pi}{2}\).

In our calculation, \(2{{\tan }^{ - 1}}\frac{1}{5} = {{\tan }^{ - 1}}\frac{5}{12}\). The value \(\frac{5}{12}\) is positive, so \({{\tan }^{ - 1}}\frac{5}{12}\) is in the range \(\left(0, \frac{\pi}{2}\right)\). The angle \(\frac{\pi}{4}\) is also in this range. When we subtract \(\frac{\pi}{4}\) from \({{\tan }^{ - 1}}\frac{5}{12}\), the result could be positive or negative, depending on which angle is larger. Since \(\frac{5}{12} < 1\), \({{\tan }^{ - 1}}\frac{5}{12} < {{\tan }^{ - 1}}1 = \frac{\pi}{4}\). Therefore, \({{\tan }^{ - 1}}\frac{5}{12} - \frac{\pi}{4}\) will be a negative angle, and its tangent will be negative, which aligns with our result \(-\frac{7}{17}\).

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Important Questions from Inverse Trigonometric Functions

  1. What is \(1+\sin ^2\left(\cos ^{-1}\left(\frac{3}{\sqrt{17}}\right)\right)\) equal to ?

  2. What is 2 cot \(\left(\frac{1}{2} \cos ^{-1} \frac{\sqrt{5}}{3}\right)\) equal to ?

  3. If sec-1 p - cosec-1q = 0, where p > 0, q > 0; then what is the value of p-2 + q-2 ?

  4. Consider the following statements:

    1. There exists \({\rm{\theta }} \in \left( { - \frac{{\rm{\pi }}}{2},\frac{{\rm{\pi }}}{2}} \right)\) for which tan -1 (tan θ) ≠ θ

    2. \({\sin ^{ - 1}}\left( {\frac{1}{3}} \right) - {\sin ^{ - 1}}\left( {\frac{1}{5}} \right) = {\sin ^{ - 1}}\left( {\frac{{2\sqrt 2 \left( {\sqrt 3 - 1} \right)}}{{15}}} \right)\)

    Which of the above statements is/are correct?

  5. Consider the following statements:

    1. \({\tan ^{ - 1}}{\rm{x}} + {\tan ^{ - 1}}\left( {\frac{1}{{\rm{x}}}} \right) = {\rm{\pi }}\)

    2. There exist x, y ∈ [-1, 1], where x ≠ y such that sin -1 x + cos -1 \({\rm{y}} = \frac{{\rm{\pi }}}{2}\)

    Which of the above statements is/are correct?
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