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