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STD 11 - 3. trignometrical ratios,functions and identities — Maths STD 11 Science — Question

CBSE BoardEnglish MediumSTD 11 ScienceMathsSTD 11 - 3. trignometrical ratios,functions and identities1 Mark
MCQ
$\frac{{\tan A + \sec A - 1}}{{\tan A - \sec A + 1}} = $
  • A
    $\frac{{1 - \sin A}}{{\cos A}}$
  • B
    $\frac{{1 - \cos A}}{{\sin A}}$
  • ✓
    $\frac{{1 + \sin A}}{{\cos A}}$
  • D
    $\frac{{1 + \cos A}}{{\sin A}}$
✓

Answer

Correct option: C.
$\frac{{1 + \sin A}}{{\cos A}}$
c
(c) $\frac{{\tan A + \sec A - 1}}{{\tan A - \sec A + 1}}$

$ = \frac{{\sin A - \cos A + 1}}{{\sin A - 1 + \cos A}} $

$= \frac{{\sin A + (1 - \cos A)}}{{\sin A - (1 - \cos A)}}$

$ = \frac{{2\sin \frac{A}{2}\cos \frac{A}{2} + 2{{\sin }^2}\frac{A}{2}}}{{2\sin \frac{A}{2}\cos \frac{A}{2} - 2{{\sin }^2}\frac{A}{2}}}$

$ = \frac{{\cos \frac{A}{2} + \sin \frac{A}{2}}}{{\cos \frac{A}{2} - \sin \frac{A}{2}}} $

$= \frac{{{{\left( {\cos \frac{A}{2} + \sin \frac{A}{2}} \right)}^2}}}{{{{\cos }^2}\frac{A}{2} - {{\sin }^2}\frac{A}{2}}}$

$ = \frac{{1 + \sin A}}{{\cos A}}$.

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