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Trigonometry – II — Maths STD 11 Science — Question

Maharashtra BoardEnglish MediumSTD 11 ScienceMathsTrigonometry – II2 Marks
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)
$\frac{\tan A+\sec A-1}{\tan A-\sec A+1}=\frac{\sin A+1-\cos A}{\sin A-1+\cos A}$
$=\frac{\sin A+(1-\cos A)}{\sin A-(1-\cos A)}$
$=\frac{2 \sin \frac{A}{2} \cos \frac{A}{2}+2 \sin ^2 \frac{A}{2}}{2 \sin \frac{A}{2} \cos \frac{A}{2}-2 \sin ^2 \frac{A}{2}}$
$=\frac{\cos \frac{ A }{2}+\sin \frac{ A }{2}}{\cos \frac{A}{2}-\sin \frac{ A }{2}}$
$=\frac{\left(\cos \frac{ A }{2}+\sin \frac{ A }{2}\right)^2}{\cos ^2 \frac{A}{2}-\sin ^2 \frac{A}{2}}$
$=\frac{1+\sin A}{\cos A}$

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