Question

If $\hat a$ and $\hat b$ are unit vectors inclined at an angle $\theta$, then prove that
$\tan \dfrac{\theta }{2} = \left| {\dfrac{{\hat a - \hat b}}{{\hat a + \hat b}}} \right|$

Answer 1

Hint: To prove this type of identity you have to start from $\left| {\vec A \pm \vec B} \right| = \sqrt {{{\left| {\vec A} \right|}^2} + {{\left| {\vec B} \right|}^2} \pm 2\vec A\vec B\cos \theta }$ here and it is given a and b are unit vectors so put A=B=1 and proceed further using trigonometric results.

Complete step-by-step answer:

Using the formula
$\left| {\vec A \pm \vec B} \right| = \sqrt {{{\left| {\vec A} \right|}^2} + {{\left| {\vec B} \right|}^2} \pm 2\vec A\vec B\cos \theta }$
Put A=B=1 because of unit vectors.
$\left| {\hat a + \hat b} \right| = \sqrt {1 + 1 + 2\cos \theta } = \sqrt {2\left( {1 + \cos \theta } \right)} = \sqrt {4{{\cos }^2}\dfrac{\theta }{2}}$ $\left( {\because \left( {1 + \cos \theta = 2{{\cos }^2}\dfrac{\theta }{2}} \right)} \right)$
$\left| {\hat a - \hat b} \right| = \sqrt {1 + 1 - 2\cos \theta } = \sqrt {2\left( {1 - \cos \theta } \right)} = \sqrt {4{{\sin }^2}\dfrac{\theta }{2}}$ $\left( {\because \left( {1 + \sin \theta = 2{{\sin }^2}\dfrac{\theta }{2}} \right)} \right)$
So we have to find
$\dfrac{{\left| {\hat a - \hat b} \right|}}{{\left| {\hat a + \hat b} \right|}} = \dfrac{{\sqrt {4{{\sin }^2}\dfrac{\theta }{2}} }}{{\sqrt {4{{\cos }^2}\dfrac{\theta }{2}} }} = \tan \dfrac{\theta }{2}$
Hence proved.

Note: Whenever you get these types of questions the key concept of solving is you have to proceed from that result which is given in hint and use what is given in question and then use trigonometric results like $\left( {1 + \cos \theta = 2{{\cos }^2}\dfrac{\theta }{2}} \right)$ to proceed further and use basic math to get an answer.