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Trigonometry – II — Maths STD 11 Science — Question

Maharashtra BoardEnglish MediumSTD 11 ScienceMathsTrigonometry – II2 Marks
MCQ
$\frac{\sin (B+A)+\cos (B-A)}{\sin (B-A)+\cos (B+A)}=$
  • A
    $\frac{\cos B +\sin B }{\cos B -\sin B }$
  • ✓
    $\frac{\cos A+\sin A}{\cos A-\sin A}$
  • C
    $\frac{\cos A-\sin A}{\cos A+\sin A}$
  • D
    $\frac{\cos B-\sin B}{\cos B+\sin B}$
✓

Answer

Correct option: B.
$\frac{\cos A+\sin A}{\cos A-\sin A}$
(B)
$\frac{\sin ( B + A )+\cos ( B - A )}{\sin ( B - A )+\cos ( B + A )}$
$=\frac{\sin ( B + A )+\sin \left\{\left(90^{\circ}-( B - A )\right\}\right.}{\sin ( B - A )+\sin \left\{\left(90^{\circ}-( A + B )\right\}\right.}$
$=\frac{2 \sin \left(A+45^{\circ}\right) \cos \left(45^{\circ}- B \right)}{2 \sin \left(45^{\circ}- A \right) \cos \left(45^{\circ}- B \right)}$
$=\frac{\sin \left(A+45^{\circ}\right)}{\sin \left(45^{\circ}-A\right)}=\frac{\cos A+\sin A}{\cos A-\sin A}$

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