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If $\mathbf{a}$ and $\mathbf{b}$ are two non-collinear vectors and the vector $\mathbf{a}+\mathbf{b}$ bisects the angle between $\mathbf{a}$ and $\mathbf{b}$, then
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The correct answer is:
$|\mathbf{a}|=|\mathbf{b}|$
Here, $O A=|\mathbf{a}|$
$O D=A C=|\mathbf{b}|$
$\begin{aligned} & \text { In } \triangle A B C, A B=|\mathbf{b}| \cos 2 \theta \\ & \text { and } B C=|\mathbf{b}| \sin 2 \theta \\ & \text { In } \triangle O B C \text {, }\end{aligned}$
$\begin{aligned} & \tan \theta=\frac{b \sin 2 \theta}{a+b \cos 2 \theta} \\ & \frac{2 b \tan \theta}{a\left(1+\tan ^2 \theta\right)+b\left(1-\tan ^2 \theta\right)}\end{aligned}$
$\begin{aligned} & \Rightarrow \quad\left[a+b+(a-b) \tan ^2 \theta-2 b\right] \tan \theta=0 \\ & \Rightarrow \quad(a-b)\left(1+\tan ^2 \theta\right) \tan \theta=0 \\ & \Rightarrow \quad(a-b) \sec ^2 \theta \tan \theta=0 \\ & \Rightarrow \quad a-b=0 \Rightarrow|\mathbf{a}|=|\mathbf{b}|\end{aligned}$
$O D=A C=|\mathbf{b}|$
$\begin{aligned} & \text { In } \triangle A B C, A B=|\mathbf{b}| \cos 2 \theta \\ & \text { and } B C=|\mathbf{b}| \sin 2 \theta \\ & \text { In } \triangle O B C \text {, }\end{aligned}$
$\begin{aligned} & \tan \theta=\frac{b \sin 2 \theta}{a+b \cos 2 \theta} \\ & \frac{2 b \tan \theta}{a\left(1+\tan ^2 \theta\right)+b\left(1-\tan ^2 \theta\right)}\end{aligned}$
$\begin{aligned} & \Rightarrow \quad\left[a+b+(a-b) \tan ^2 \theta-2 b\right] \tan \theta=0 \\ & \Rightarrow \quad(a-b)\left(1+\tan ^2 \theta\right) \tan \theta=0 \\ & \Rightarrow \quad(a-b) \sec ^2 \theta \tan \theta=0 \\ & \Rightarrow \quad a-b=0 \Rightarrow|\mathbf{a}|=|\mathbf{b}|\end{aligned}$
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