Question

The value of \[\dfrac{{\cot x - \tan x}}{{\cot 2x}}\] is
a) 1
b) 2
c) -1
d) 4

Answer 1

Hint: Here the question is related to trigonometry. Firstly we write the given term by using the definition of the trigonometric ratio. As we know that every trigonometric ratio is interlinked to each other. Then by using the double angle formula we are obtaining the value of \[\dfrac{{\cot x - \tan x}}{{\cot 2x}}\].

Complete step by step answer:
The sine, cosine, tangent, cosecant, secant and cotangent are the trigonometry ratios of trigonometry. It is abbreviated as sin, cos, tan, cosec, sec and cot.
We have totally 6 trigonometric ratios in the trigonometry. The 4 trigonometry ratios are depended on the other 2 trigonometric ratios.
The trigonometric ratios are sine, cosine, tangent, cosecant, secant and cotangent.
The tangent trigonometric ratio is the ratio of the sine and cosine trigonometric ratio.
\[ \Rightarrow \tan x = \dfrac{{\sin x}}{{\cos x}}\]
The cosecant trigonometric ratio is the reciprocal of the sine trigonometric ratio.
\[ \Rightarrow \csc x = \dfrac{1}{{\sin x}}\]
The secant trigonometric ratio is the reciprocal of the cosine trigonometric ratios.
\[ \Rightarrow \sec x = \dfrac{1}{{\cos x}}\]
The cotangent trigonometric ratio is the reciprocal of the tangent trigonometric ratio.
\[ \Rightarrow \cot x = \dfrac{1}{{\tan x}}\]
Now consider the given question
\[ \Rightarrow \dfrac{{\cot x - \sin x}}{{\cot 2x}}\]
This can be written as
\[ \Rightarrow \dfrac{{\dfrac{{\cos x}}{{\sin x}} - \dfrac{{\sin x}}{{\cos x}}}}{{\dfrac{{\cos 2x}}{{\sin 2x}}}}\]
On simplifying the numerator term we have
\[ \Rightarrow \dfrac{{\dfrac{{{{\cos }^2}x - {{\sin }^2}x}}{{\sin x\cos x}}}}{{\dfrac{{\cos 2x}}{{\sin 2x}}}}\]
As we know the double angle for the trigonometric ratio \[\cos 2x = {\cos ^2}x - {\sin ^2}x\] and \[\sin 2x = 2\sin x\cos x\]
On considering the above formulas the above inequality is written as
\[ \Rightarrow \dfrac{{\dfrac{{{{\cos }^2}x - {{\sin }^2}x}}{{\sin x\cos x}}}}{{\dfrac{{{{\cos }^2}x - \sin^2 x}}{{2\sin x\cos x}}}}\]
On taking the reciprocal we have
\[ \Rightarrow \dfrac{{{{\cos }^2}x - {{\sin }^2}x}}{{\sin x\cos x}} \times \dfrac{{2\sin x\cos x}}{{{{\cos }^2}x - {{\sin }^2}x}}\]
On cancelling the terms we have
\[ \Rightarrow 2\]
Therefore, the value of \[\dfrac{{\cot x - \tan x}}{{\cot 2x}}\] is 2. Hence the option (b) is the correct option.

Note:
When the trigonometric ratio is in the form of double angle then we have to know the formulas, they are \[\cos 2x = {\cos ^2}x - {\sin ^2}x = 1 - 2{\sin ^2}x = 2{\cos ^2}x - 1 = \dfrac{{1 - {{\tan }^2}x}}{{1 + {{\tan }^2}x}}\], \[\sin 2x = 2\sin x\cos x\] and \[\tan 2x = \dfrac{{2\tan x}}{{1 - {{\tan }^2}x}}\]. The definition of the trigonometric ratios are applied to these kinds of problems.