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STD 12 - 2. inverse trigonometric function — Mathematics STD 12 Science — Question

CBSE BoardEnglish MediumSTD 12 ScienceMathematicsSTD 12 - 2. inverse trigonometric function1 Mark
MCQ
If ${a_{1,}}{a_2},{a_3}.....,{a_n}$ is an $A.P.$ with common difference d then $\tan \left[ {{{\tan }^{ - 1}}\left( {\frac{d}{{1 + {a_1}{a_2}}}} \right) + {{\tan }^{ - 1}}\left( {\frac{d}{{1 + {a_2}{a_3}}}} \right) + ...} \right.$ $\left. { + {{\tan }^{ - 1}}\left( {\frac{d}{{1 + {a_{n - 1}}{a_n}}}} \right)} \right] = $
  • A
    $\frac{{(n - 1)d}}{{{a_1} + {a_n}}}$
  • ✓
    $\frac{{(n - 1)d}}{{1 + {a_1}{a_n}}}$
  • C
    $\frac{{nd}}{{1 + {a_1}{a_n}}}$
  • D
    $\frac{{{a_n} - {a_1}}}{{{a_n} + {a_1}}}$
✓

Answer

Correct option: B.
$\frac{{(n - 1)d}}{{1 + {a_1}{a_n}}}$
b
(b) We have ${\tan ^{ - 1}}\left( {\frac{d}{{1 + {a_1}{a_2}}}} \right) + {\tan ^{ - 1}}\left( {\frac{d}{{1 + {a_2}{a_3}}}} \right) + .....$
$..... + {\tan ^{ - 1}}\left( {\frac{d}{{1 + {a_{n - 1}}{a_n}}}} \right)$
$ = {\tan ^{ - 1}}\left( {\frac{{{a_2} - {a_1}}}{{1 + {a_1}{a_2}}}} \right) + {\tan ^{ - 1}}\left( {\frac{{{a_3} - {a_2}}}{{1 + {a_2}{a_3}}}} \right) + .....$
$..... + {\tan ^{ - 1}}\left( {\frac{{{a_n} - {a_{n - 1}}}}{{1 + {a_{n - 1}}{a_n}}}} \right)$
$ = ({\tan ^{ - 1}}{a_2} - {\tan ^{ - 1}}{a_1}) + ({\tan ^{ - 1}}{a_3} - {\tan ^{ - 1}}{a_2}) + ....$
$..... + ({\tan ^{ - 1}}{a_n} - {\tan ^{ - 1}}{a_{n - 1}})$
$ = {\tan ^{ - 1}}{a_n} - {\tan ^{ - 1}}{a_1} = {\tan ^{ - 1}}\left( {\frac{{{a_n} - {a_1}}}{{1 + {a_n}{a_1}}}} \right)$
$ = {\tan ^{ - 1}}\left( {\frac{{(n - 1)d}}{{1 + {a_1}{a_n}}}} \right)$.

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