If \({\sin ^{ - 1}}\frac{{2p}}{{1 + p2}} - {\cos ^{ - 1}}\frac{{1 - {q^2}}}{{1 + {q^2}}} = {\tan ^{ - 1}}\frac{{2x}}{{1 - {x^2}}}\) , then what is x equal to?
Understanding the Inverse Trigonometry Problem
The question asks us to find the value of \(x\) given an equation involving inverse trigonometric functions. The equation is:
\[ {\sin ^{ - 1}}\frac{{2p}}{{1 + p2}} - {\cos ^{ - 1}}\frac{{1 - {q^2}}}{{1 + {q^2}}} = {\tan ^{ - 1}}\frac{{2x}}{{1 - {x^2}}} \]
To solve this, we need to use standard identities relating inverse sine, cosine, and tangent functions, specifically those that convert them into the form \(2\tan^{-1}y\).
Applying Inverse Trigonometric Identities
We will use the following key identities:
The identity for inverse sine: \(2\tan^{-1}y = \sin^{-1}\frac{2y}{1+y^2}\)
The identity for inverse cosine: \(2\tan^{-1}y = \cos^{-1}\frac{1-y^2}{1+y^2}\)
The identity for inverse tangent: \(2\tan^{-1}y = \tan^{-1}\frac{2y}{1-y^2}\)
Using these identities, we can rewrite each term in the given equation:
The first term, \({\sin ^{ - 1}}\frac{{2p}}{{1 + p2}}\), matches the form \({\sin ^{ - 1}}\frac{{2y}}{{1 + y^2}}\) with \(y=p\). So, \({\sin ^{ - 1}}\frac{{2p}}{{1 + p2}} = 2\tan^{-1}p\).
The second term, \({\cos ^{ - 1}}\frac{{1 - {q^2}}}{{1 + {q^2}}}\), matches the form \({\cos ^{ - 1}}\frac{{1 - y^2}}{{1 + y^2}}\) with \(y=q\). So, \({\cos ^{ - 1}}\frac{{1 - {q^2}}}{{1 + {q^2}}} = 2\tan^{-1}q\).
The third term, \({\tan ^{ - 1}}\frac{{2x}}{{1 - {x^2}}}\), matches the form \({\tan ^{ - 1}}\frac{{2y}}{{1 - y^2}}\) with \(y=x\). So, \({\tan ^{ - 1}}\frac{{2x}}{{1 - {x^2}}} = 2\tan^{-1}x\).
Substituting Identities into the Equation
Now, substitute these equivalent forms back into the original equation:
\[ (2\tan^{-1}p) - (2\tan^{-1}q) = (2\tan^{-1}x) \]
Simplifying and Solving for x
We can divide the entire equation by 2:
\[ \tan^{-1}p - \tan^{-1}q = \tan^{-1}x \]
Next, we use the identity for the difference of two inverse tangent functions:
\[ \tan^{-1}A - \tan^{-1}B = \tan^{-1}\frac{A-B}{1+AB} \]
Applying this identity to the left side of our simplified equation (with \(A=p\) and \(B=q\)):
\[ \tan^{-1}\frac{p-q}{1+(p)(q)} = \tan^{-1}x \]
For the equality of the inverse tangent functions to hold, their arguments must be equal:
\[ \frac{p-q}{1+pq} = x \]
Therefore, the value of \(x\) is \(\frac{p-q}{1+pq}\).
Conclusion
By using the standard identities for inverse trigonometric functions, we transformed the given equation into a simpler form involving only \(\tan^{-1}\) terms, allowing us to isolate and find the value of \(x\).
The calculated value of \(x\) is \(\frac{{p - q}}{{1 + pq}}\).
| Step | Action | Result |
|---|---|---|
| 1 | Identify the structure of the terms in the equation. | Match forms like \(\sin^{-1}\frac{2p}{1+p^2}\), \(\cos^{-1}\frac{1-q^2}{1+q^2}\), \(\tan^{-1}\frac{2x}{1-x^2}\). |
| 2 | Apply the identity \(2\tan^{-1}y\) to each term. | \(2\tan^{-1}p - 2\tan^{-1}q = 2\tan^{-1}x\) |
| 3 | Simplify the equation. | \(\tan^{-1}p - \tan^{-1}q = \tan^{-1}x\) |
| 4 | Apply the difference identity for \(\tan^{-1}\). | \(\tan^{-1}\frac{p-q}{1+pq} = \tan^{-1}x\) |
| 5 | Equate arguments to find x. | \(x = \frac{p-q}{1+pq}\) |
Revision Table: Inverse Trigonometry Formulas
| Identity | Formula |
|---|---|
| Sine to Tangent Inverse | \(2\tan^{-1}y = \sin^{-1}\frac{2y}{1+y^2}\) |
| Cosine to Tangent Inverse | \(2\tan^{-1}y = \cos^{-1}\frac{1-y^2}{1+y^2}\) |
| Tangent Inverse Addition | \(\tan^{-1}A + \tan^{-1}B = \tan^{-1}\frac{A+B}{1-AB}\) |
| Tangent Inverse Subtraction | \(\tan^{-1}A - \tan^{-1}B = \tan^{-1}\frac{A-B}{1+AB}\) |
Additional Information: Conditions for Identities
It's important to note that the identities used, such as \(2\tan^{-1}y = \sin^{-1}\frac{2y}{1+y^2}\) and \(2\tan^{-1}y = \cos^{-1}\frac{1-y^2}{1+y^2}\), have specific conditions on \(y\) for the equality to hold exactly as written (usually \(|y| \le 1\)). Similarly, the identity \(2\tan^{-1}y = \tan^{-1}\frac{2y}{1-y^2}\) requires \(|y| < 1\). The sum/difference formulas for \(\tan^{-1}\) also have conditions related to \(AB < 1\) for the principal value. In typical problems like this from competitive exams, it's often assumed that the values of p, q, and x fall within the ranges where these principal value identities are valid, unless specified otherwise.
Understanding these conditions is crucial for a complete grasp of inverse trigonometric functions, but for solving this particular problem based on the provided options, using the direct identities is sufficient.
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