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Question: $\frac{sin(B + A) + cos(B - A)}{sin(B - A) + cos(B + A)}$ = [IIT JEE]...

sin(B+A)+cos(B−A)sin(B−A)+cos(B+A)\frac{sin(B + A) + cos(B - A)}{sin(B - A) + cos(B + A)} = [IIT JEE]

A

cosB+sinBcosB−sinB\frac{cos B + sin B}{cos B - sin B}

B

cosA+sinAcosA−sinA\frac{cos A + sin A}{cos A - sin A}

C

cosA−sinAcosA+sinA\frac{cos A - sin A}{cos A + sin A}

D

None of these

Answer

(b) cosA+sinAcosA−sinA\frac{cos A + sin A}{cos A - sin A}

Explanation

Solution

To simplify the expression sin⁡(B+A)+cos⁡(B−A)sin⁡(B−A)+cos⁡(B+A)\frac{\sin(B + A) + \cos(B - A)}{\sin(B - A) + \cos(B + A)}, we use the sum and difference identities for sine and cosine:

  1. sin⁡(X+Y)=sin⁡Xcos⁡Y+cos⁡Xsin⁡Y\sin(X+Y) = \sin X \cos Y + \cos X \sin Y
  2. sin⁡(X−Y)=sin⁡Xcos⁡Y−cos⁡Xsin⁡Y\sin(X-Y) = \sin X \cos Y - \cos X \sin Y
  3. cos⁡(X+Y)=cos⁡Xcos⁡Y−sin⁡Xsin⁡Y\cos(X+Y) = \cos X \cos Y - \sin X \sin Y
  4. cos⁡(X−Y)=cos⁡Xcos⁡Y+sin⁡Xsin⁡Y\cos(X-Y) = \cos X \cos Y + \sin X \sin Y

Expanding the numerator: sin⁡(B+A)+cos⁡(B−A)=(sin⁡Bcos⁡A+cos⁡Bsin⁡A)+(cos⁡Bcos⁡A+sin⁡Bsin⁡A)=(sin⁡B+cos⁡B)(cos⁡A+sin⁡A)\sin(B + A) + \cos(B - A) = (\sin B \cos A + \cos B \sin A) + (\cos B \cos A + \sin B \sin A) = (\sin B + \cos B)(\cos A + \sin A)

Expanding the denominator: sin⁡(B−A)+cos⁡(B+A)=(sin⁡Bcos⁡A−cos⁡Bsin⁡A)+(cos⁡Bcos⁡A−sin⁡Bsin⁡A)=(sin⁡B+cos⁡B)(cos⁡A−sin⁡A)\sin(B - A) + \cos(B + A) = (\sin B \cos A - \cos B \sin A) + (\cos B \cos A - \sin B \sin A) = (\sin B + \cos B)(\cos A - \sin A)

Therefore, the expression simplifies to: (sin⁡B+cos⁡B)(cos⁡A+sin⁡A)(sin⁡B+cos⁡B)(cos⁡A−sin⁡A)=cos⁡A+sin⁡Acos⁡A−sin⁡A\frac{(\sin B + \cos B)(\cos A + \sin A)}{(\sin B + \cos B)(\cos A - \sin A)} = \frac{\cos A + \sin A}{\cos A - \sin A}