The angle between two vectors A→ and B→ is θ. The resultant of these vectors R→ makes an angle of

tan⁡(θ/2)=Bsin⁡θA+Bcos⁡θ⇒sin⁡(θ/2)cos⁡(θ/2)=2Bsin⁡(θ/2)cos⁡(θ/2)A+B2cos2⁡(θ/2)−1⇒A+2Bcos2⁡(θ/2)−B=2Bcos2⁡(θ/2)A=B

The angle between two vectors A→ and B→ is θ. The resultant of these vectors R→ makes an angle of

tan⁡(θ/2)=Bsin⁡θA+Bcos⁡θ⇒sin⁡(θ/2)cos⁡(θ/2)=2Bsin⁡(θ/2)cos⁡(θ/2)A+B2cos2⁡(θ/2)−1⇒A+2Bcos2⁡(θ/2)−B=2Bcos2⁡(θ/2)A=B

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The angle between two vectors $\vec{A}$ and $\vec{B}$ is $θ$ . The resultant of these vectors $\vec{R}$ makes an angle of

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A = 2B

A = B/2

A = B

AB = 1

answer is C.

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Detailed Solution

$\begin{array}{l} \tan (θ / 2) = \frac{Bsin θ}{A + Bcos θ} \\ \Rightarrow \frac{\sin (θ / 2)}{\cos (θ / 2)} = \frac{2 Bsin (θ / 2) \cos (θ / 2)}{A + B [2 \cos^{2} (θ / 2) - 1]} \\ \Rightarrow A + 2 {Bcos}^{2} (θ / 2) - B = 2 {Bcos}^{2} (θ / 2) \\ A = B \end{array}$