# Prove that:

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

Last updated date: 26th Mar 2023

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Answer

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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.

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.

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