Question

If $ \cot x = 2 $ , find the value of $ \dfrac{{\left( {2 + 2\sin x} \right)\left( {1 - \sin x} \right)}}{{\left( {1 + \cos x} \right)\left( {2 - 2\cos x} \right)}} $ ?

Answer 1

Hint: The given question is related to the concept of trigonometric functions. $ \cot x $ is a trigonometric function and is opposite of another trigonometric function, $ \tan x $ . We know that the value of x remains between $ {0^ \circ } $ and $ {360^ \circ } $ . The domain of both $ \sin x $ and $ \cos x $ is $ \left( { - \infty ,\infty } \right) $ and range is between $ \left[ { - 1,1} \right] $ . Here, in this question we have to find the value of a given function. We will use trigonometric identities and some algebraic identities to solve the given problem.

Complete step by step solution:
Given is $ \cot x = 2 $
According to the question, let us try to simplify the function and solve it.
First, we take out $ 2 $ as it is common in both numerator and denominator.
$
   \Rightarrow \dfrac{{2\left( {1 + \sin x} \right)\left( {1 - \sin x} \right)}}{{2\left( {1 - \cos x} \right)\left( {1 + \cos x} \right)}} \\
   \Rightarrow \dfrac{{\left( {1 + \sin x} \right)\left( {1 - \sin x} \right)}}{{\left( {1 - \cos x} \right)\left( {1 + \cos x} \right)}} \\
$
Applying the algebraic identity $ \left( {a + b} \right)\left( {a - b} \right) = {a^2} - {b^2} $ , we get;
$ \Rightarrow \dfrac{{1 - {{\sin }^2}x}}{{1 - {{\cos }^2}x}} $
We know that $ {\sin ^2}x + {\cos ^2}x = 1 $ , so using the same we say that
$ 1 - {\sin ^2}x = {\cos ^2}x $
$ 1 - {\cos ^2}x = {\sin ^2}x $
Therefore, we get,
$ \Rightarrow \dfrac{{{{\cos }^2}x}}{{{{\sin }^2}x}} $
We know $ \tan \theta = \dfrac{{\sin \theta }}{{\cos \theta }} $ and $ \cot \theta $ is the reciprocal of $ \tan \theta $ i.e., $ \cot \theta = \dfrac{{\cos \theta }}{{\sin \theta }} $ . So, we get;
$ \Rightarrow {\cot ^2}\theta $
We are given the value of $ \cot x $ as $ 2 $ . Using the same, we get,
$
   \Rightarrow {\left( 2 \right)^2} \\
   \Rightarrow 4 \\
$
Hence, the value of $ \dfrac{{\left( {2 + 2\sin x} \right)\left( {1 - \sin x} \right)}}{{\left( {1 + \cos x} \right)\left( {2 - 2\cos x} \right)}} $ is $ 4 $ .

Note: Here, in this question we used trigonometric identity as well as algebraic identity and we could easily find the value. It is highly recommended to keep all the trigonometric identities in mind while solving trigonometry questions as without using these identities, solving a question would be difficult and take a lot of time. Students should also keep in their mind the basic algebraic identities as they are used a lot. Students can find a lot of similar questions in their NCERT textbook. They should practice them for more clarity.