MCQOPTIONS
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MCQOPTIONS
| 1. |
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 |
| A. | \(\frac{{a + b}}{{1 + ab}}\) |
| B. | \(\frac{{a - b}}{{1 + ab}}\) |
| C. | \(\frac{{a - b}}{{1 - ab}}\) |
| D. | None of the above |
| Answer» C. \(\frac{{a - b}}{{1 - ab}}\) | |