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Question: If n is any integer, then the general solution of the equation \(R_{1}\) is....

If n is any integer, then the general solution of the equation R1R_{1} is.

A

4sin⁡4θ=−24\sin 4\theta = - 2 or sin⁡4θ=−12\sin 4\theta = \frac{- 1}{2}

B

4θ=7π64\theta = \frac{7\pi}{6}

C

11π6\frac{11\pi}{6} or 0<4θ<2π0 < 4\theta < 2\pi

D

∴\therefore or 0<θ<π20 < \theta < \frac{\pi}{2}

Answer

11π6\frac{11\pi}{6} or 0<4θ<2π0 < 4\theta < 2\pi

Explanation

Solution

Given equation is,1−tan⁡A2tan⁡B2=cos⁡A2cos⁡B2−sin⁡A2sin⁡B2cos⁡A2cos⁡B21 - \tan\frac{A}{2}\tan\frac{B}{2} = \frac{\cos\frac{A}{2}\cos\frac{B}{2} - \sin\frac{A}{2}\sin\frac{B}{2}}{\cos\frac{A}{2}\cos\frac{B}{2}}

Dividing equation by =cos⁡(A2+B2)cos⁡A2cos⁡B2=sin⁡C2cos⁡A2cos⁡B2= \frac{\cos\left( \frac{A}{2} + \frac{B}{2} \right)}{\cos\frac{A}{2}\cos\frac{B}{2}} = \frac{\sin\frac{C}{2}}{\cos\frac{A}{2}\cos\frac{B}{2}},

b2cos⁡2A−a2cos⁡2Bb^{2}\cos 2A - a^{2}\cos 2B

=b2(1−2sin⁡2A)−a2(1−2sin⁡2B)= b^{2}(1 - 2\sin^{2}A) - a^{2}(1 - 2\sin^{2}B). Hence, =b2−a2−2(b2sin⁡2A−a2sin⁡2B)=b2−a2= b^{2} - a^{2} - 2(b^{2}\sin^{2}A - a^{2}\sin^{2}B) = b^{2} - a^{2}

asin⁡(B−C)+bsin⁡(C−A)+csin⁡(A−B)a\sin(B - C) + b\sin(C - A) + c\sin(A - B)

or=k(Σsin⁡Asin⁡(B−C)= k(\Sigma\sin A\sin(B - C).