Question

Prove that $\sqrt {\dfrac{{1 + \cos \theta }}{{1 - \cos \theta }}} + \sqrt {\dfrac{{1 - \cos \theta }}{{1 + \cos \theta }}} = 2cosec\theta $

Answer 1

Hint - Consider LHS part and start simplification and after that use trigonometric identity ${\sin ^2}\theta + {\cos ^2}\theta = 1 \Rightarrow 1 - {\cos ^2}\theta = {\sin ^2}\theta $.
 We have to prove that $\sqrt {\dfrac{{1 + \cos \theta }}{{1 - \cos \theta }}} + \sqrt {\dfrac{{1 - \cos \theta }}{{1 + \cos \theta }}} = 2cosec\theta $
Taking LHS
$\sqrt {\dfrac{{1 + \cos \theta }}{{1 - \cos \theta }}} + \sqrt {\dfrac{{1 - \cos \theta }}{{1 + \cos \theta }}} $
Now let’s take the LCM to simplify above we have
$\dfrac{{{{\left( {\sqrt {1 + \cos \theta } } \right)}^2} + {{\left( {\sqrt {1 - \cos \theta } } \right)}^2}}}{{\sqrt {1 + \cos \theta } \sqrt {1 - \cos \theta } }}$
Now using ${\left( {\sqrt a } \right)^2} = a$ we have
$
  \dfrac{{1 + \cos \theta + 1 - \cos \theta }}{{\sqrt {1 + \cos \theta } \sqrt {1 - \cos \theta } }} \\
   \Rightarrow \dfrac{2}{{\sqrt {1 + \cos \theta } \sqrt {1 - \cos \theta } }} \\
   \Rightarrow \dfrac{2}{{\sqrt {\left( {1 - \cos \theta } \right)\left( {1 + \cos \theta } \right)} }} \\
 $
Now in denominator part we have $\left( {a - b} \right)\left( {a + b} \right) = {a^2} - {b^2}$
So we get
$\dfrac{2}{{\sqrt {1 - {{\cos }^2}\theta } }}$
Using the trigonometric identity ${\sin ^2}\theta + {\cos ^2}\theta = 1 \Rightarrow 1 - {\cos ^2}\theta = {\sin ^2}\theta $
We get
$
  \dfrac{2}{{\sqrt {{{\sin }^2}\theta } }} \\
   \Rightarrow \dfrac{2}{{\sin \theta }} \\
 $
Now we know that $\sin \theta = \dfrac{1}{{cosec\theta }}{\text{ or cosec}}\theta {\text{ = }}\dfrac{1}{{\sin \theta }}$
So the above get reduced to
$2cosec\theta $ =RHS
Hence LHS=RHS
 Thus proved
Note – Whenever we come across such problems the basics that we need to follow is that we always start from one side of the given equation and simplify that part using various trigonometric identities . A good grasp over the trigonometric identities is appreciated to reach to the proof.