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) $\frac{{\tan A + \sec A - 1}}{{\tan A - \sec A + 1}}$
(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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