Question

# Prove that:$\left( {1 + {{\cot }^2}\theta } \right).\left( {1 - \cos \theta } \right).\left( {1 + \cos \theta } \right) = 1$

Hint- Use the following formulae ${\text{ }}\cot \theta = \dfrac{{\cos \theta }}{{\sin \theta }},{\text{ }}\left( {a - b} \right)\left( {a + b} \right) = {a^2} - {b^2}$

We have to prove
$\left( {1 + {{\cot }^2}\theta } \right).\left( {1 - \cos \theta } \right).\left( {1 + \cos \theta } \right) = 1$
Consider L.H.S
$\left( {1 + {{\cot }^2}\theta } \right).\left( {1 - \cos \theta } \right).\left( {1 + \cos \theta } \right)$
As we know
${\text{ }}\cot \theta = \dfrac{{\cos \theta }}{{\sin \theta }},{\text{ }}\left( {a - b} \right)\left( {a + b} \right) = {a^2} - {b^2}$
So, apply these trigonometric properties in above equation
$\Rightarrow \left( {1 + {{\left( {\dfrac{{\cos \theta }}{{\sin \theta }}} \right)}^2}} \right)\left( {1 - {{\cos }^2}\theta } \right) \\ \Rightarrow \left( {\dfrac{{{{\sin }^2}\theta + {{\cos }^2}\theta }}{{{{\sin }^2}\theta }}} \right)\left( {1 - {{\cos }^2}\theta } \right) \\$
Now we know that ${\sin ^2}\theta + {\cos ^2}\theta = 1,{\text{ }}\left( {1 - {{\cos }^2}\theta } \right) = {\sin ^2}\theta$
So, apply these trigonometric properties in above equation
$\Rightarrow \left( {\dfrac{{{{\sin }^2}\theta + {{\cos }^2}\theta }}{{{{\sin }^2}\theta }}} \right)\left( {1 - {{\cos }^2}\theta } \right) = \dfrac{1}{{{{\sin }^2}\theta }}{\sin ^2}\theta \\ = 1 \\$
= R.H.S
Hence Proved

Note- In such types of questions always remember the general trigonometric identities which are stated above and using these properties simplify the given equation we will get the required answer.