Question

Value of the expression
$\left( {1 - \cos \theta } \right)\left( {1 + \cos \theta } \right)\left( {1 + {{\cot }^2}\theta } \right)$ is
A). $0$
B). $1$
C). ${\sin ^2}\theta$
D). ${\csc ^2}\theta$

Answer 1

In the given question, we have been given an expression involving the use of trigonometric functions. We have to find the value of this condensed expression. We are going to solve it by using the basic trigonometric and algebraic identities by identifying them with their resulting values. Then we are going to write their result, simplify their value and calculate the answer.

Formula used:
We are going to use the formula of difference of squares, which is:
$\left( {a + b} \right)\left( {a - b} \right) = {a^2} - {b^2}$

Complete step by step solution:
The given expression is $p = \left( {1 - \cos \theta } \right)\left( {1 + \cos \theta } \right)\left( {1 + {{\cot }^2}\theta } \right)$.
First, we are going to simplify the value of the first two brackets using the formula of difference of squares, which is:
$\left( {a + b} \right)\left( {a - b} \right) = {a^2} - {b^2}$
So, $\left( {1 - \cos \theta } \right)\left( {1 + \cos \theta } \right) = 1 - {\cos ^2}\theta$
Now, we know that ${\sin ^2}\theta + {\cos ^2}\theta = 1$.
Also, $1 + {\cot ^2}\theta = {\csc ^2}\theta$
So, $p = {\sin ^2}\theta \times {\csc ^2}\theta = {\left( {\sin \theta \times \csc \theta } \right)^2}$
We know, $\sin \theta \times \csc \theta = 1$
Hence, $p = {\left( 1 \right)^2} = 1$
Thus, the correct option is (B).

Note: In the given question, we had to find the value of a condensed expression which used some trigonometric functions. We solved it by using the basic identities of algebra and trigonometry. Then we used their result to write the value of the formulae, simplified it and found our answer. So, it is very important that we know all the basic formulae and have enough practice with them so that we can identify them and remember their results.