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Question

In the equation

\({\cos ^{ - 1}}\left( {\frac{{1 - {a^2}}}{{1 + {a^2}}}} \right) - {\cos ^{ - 1}}\left( {\frac{{1 - {b^2}}}{{1 + {b^2}}}} \right) = 2{\tan ^{ - 1}}x\)

value of x is

The correct answer is \(\frac{{a - b}}{{1 + ab}}\)

Solving the Trigonometric Equation for the Value of x

We are given the trigonometric equation:

\({\cos ^{ - 1}}\left( {\frac{{1 - {a^2}}}{{1 + {a^2}}}} \right) - {\cos ^{ - 1}}\left( {\frac{{1 - {b^2}}}{{1 + {b^2}}}} \right) = 2{\tan ^{ - 1}}x\)

Our goal is to solve this equation and find the value of x.

Using Key Inverse Trigonometric Identities

To simplify the given trigonometric equation, we can use a standard identity relating cos inverse and tan inverse functions. The relevant identity is:

\({\cos^{ - 1}}\left( {\frac{{1 - {y^2}}}{{1 + {y^2}}}} \right) = 2{\tan^{ - 1}}y\)

Assuming appropriate ranges for 'a' and 'b' where this identity holds, we can apply it to the terms in the equation:

  • The first term, \({\cos ^{ - 1}}\left( {\frac{{1 - {a^2}}}{{1 + {a^2}}}} \right)\), becomes \(2{\tan^{ - 1}}a\).
  • The second term, \({\cos ^{ - 1}}\left( {\frac{{1 - {b^2}}}{{1 + {b^2}}}} \right)\), becomes \(2{\tan^{ - 1}}b\).

Substituting these into the original trigonometric equation, we get:

\(2{\tan^{ - 1}}a - 2{\tan^{ - 1}}b = 2{\tan^{ - 1}}x\)

We can simplify this equation by dividing both sides by 2:

\({\tan^{ - 1}}a - {\tan^{ - 1}}b = {\tan^{ - 1}}x\)

Applying the Difference Identity for Tan Inverse

Now we have an equation involving the difference of two tan inverse terms. We use another key identity for inverse trigonometric functions:

\({\tan^{ - 1}}A - {\tan^{ - 1}}B = {\tan^{ - 1}}\left( {\frac{{A - B}}{{1 + AB}}} \right)\)

Applying this identity to the left side of our simplified equation (with A = a and B = b), we get:

\({\tan^{ - 1}}\left( {\frac{{a - b}}{{1 + ab}}} \right) = {\tan^{ - 1}}x\)

Finding the Value of x

Since the inverse tangent function is one-to-one, if \({\tan^{ - 1}}P = {\tan^{ - 1}}Q\), then P must equal Q. Therefore, we can equate the arguments of the tan inverse function on both sides:

\(x = \frac{{a - b}}{{1 + ab}}\)

This gives us the value of x that solves the original trigonometric equation.

This process demonstrates how to solve the equation by effectively using identities for inverse trigonometric functions and simplifying the expressions step-by-step to find the value of x.

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Important Questions from Inverse Trigonometric Functions

  1. The imaginary part of log sin (x + iy) is:

  2. The value of \({\tan ^{ - 1}}\left( {\frac{1}{2}} \right) + {\tan ^{ - 1}}\left( {\frac{1}{3}} \right)\) is

  3. The function \(f(x) = \sqrt {\cos (\sin x)} + {\sin ^{ - 1}}\left( {\frac{{1 + {x^2}}}{{2x}}} \right)\) is defined for

  4. The value of \({\cos ^{ - 1}}\left( {\cos \frac{{5\pi }}{3}} \right) + {\sin ^{ - 1}}\left( {\sin \frac{{5\pi }}{3}} \right)\) is

  5. \(\sin ^{-1} \dfrac{3}{5} - \cos ^{-1} \dfrac{12}{13}\) equals to
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