Courses
Courses for Kids
Free study material
Free LIVE classes
More
Questions & Answers
seo-qna
LIVE
Join Vedantu’s FREE Mastercalss

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

Answer
VerifiedVerified
363.3k+ views
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.
Last updated date: 16th Sep 2023
•
Total views: 363.3k
•
Views today: 3.63k