Question

Simplify the trigonometric expression \[\csc A – 2\cot 2A \cos A\].
A. \[2\sin A\]
B. \[\sec A\]
C. \[2\cos A \cot A\]
D. None of these

Answer 1

Hint: First, simplify the given expression by using the trigonometric ratio \[\cot x = \dfrac{1}{{\tan x }}\]. Then apply the trigonometric identity \[\tan 2x = \dfrac{{2\tan x}}{{1 - \tan^{2}x}}\] . Further simplify the expression by using the trigonometric ratio \[\tan x = \dfrac{{\sin x }}{{\cos x }}\]. In the end, apply the trigonometric identity \[\cos^{2}x + \sin^{2}x = 1\] and simplify the expression to reach the required answer.

Formula used:
1. \[\tan 2x = \dfrac{{2\tan x}}{{1 - \tan^{2}x}}\]
2. \[\cos^{2}x + \sin^{2}x = 1\]
3. \[\tan x = \dfrac{{\sin x }}{{\cos x }}\]
4. \[\cot x = \dfrac{1}{{\tan x }}\]
5. \[\csc x = \dfrac{1}{{\sin x }}\]

Complete step by step solution:
The given trigonometric expression is \[\csc A – 2\cot 2A \cos A\].
Let’s simplify the above expression.
Apply the trigonometric ratio \[\cot x = \dfrac{1}{{\tan x }}\].
\[\csc A – 2\cot 2A \cos A = \csc A - 2\left( {\dfrac{1}{{\tan 2A}}} \right) \cos A\]
Use the double angle formula \[\tan 2x = \dfrac{{2\tan x}}{{1 - \tan^{2}x}}\].
\[\csc A – 2\cot 2A \cos A = \csc A - 2\left( {\dfrac{1}{{\dfrac{{2\tan A}}{{1 - \tan^{2}A}}}}} \right) \cos A\]
\[ \Rightarrow \csc A – 2\cot 2A \cos A = \csc A - \left( {\dfrac{{2\cos A}}{{2\tan A}}} \right) \left( {1 - \tan^{2}A} \right)\]
Now apply the trigonometric ratio \[\tan x = \dfrac{{\sin x }}{{\cos x }}\].
\[\csc A - 2\cot 2A \cos A = \csc A - \left( {\dfrac{{\cos A}}{{\dfrac{{\sin A}}{{\cos A}}}}} \right) \left( {1 - \dfrac{{\sin^{2}A}}{{\cos^{2}A}}} \right)\]
Simplify the right-hand side of the above equation.
\[\csc A - 2\cot 2A \cos A = \csc A - \left( {\dfrac{{\cos^{2}A}}{{\sin A}}} \right) \left( {\dfrac{{\cos^{2}A - \sin^{2}A}}{{\cos^{2}A}}} \right)\]
\[ \Rightarrow \csc A - 2\cot 2A \cos A = \csc A - \left( {\dfrac{1}{{\sin A}}} \right) \left( {\cos^{2}A - \sin^{2}A} \right)\]
Use the trigonometric ratio \[\csc x = \dfrac{1}{{\sin x }}\].
\[csc A - 2\cot 2A \cos A = \csc A - \left( {\csc A} \right) \left( {\cos^{2}A - \sin^{2}A} \right)\]
Factor out the common term from the right-hand side.
\[\csc A - 2\cot 2A \cos A = \csc A \left( {1 - \left( {\cos^{2}A - \sin^{2}A} \right)} \right)\]
\[ \Rightarrow \csc A - 2\cot 2A \cos A = \csc A \left( {1 - \cos^{2}A + \sin^{2}A} \right)\]

Now apply the trigonometric identity \[\cos^{2}x + \sin^{2}x = 1\].
We get,
\[\csc A - 2\cot 2A \cos A = \csc A \left( {\cos^{2}A + \sin^{2}A - \cos^{2}A + \sin^{2}A} \right)\]
\[ \Rightarrow \csc A - 2\cot 2A \cos A = \csc A \left( {\sin^{2}A + \sin^{2}A} \right)\]
\[ \Rightarrow \csc A - 2\cot 2A \cos A = \csc A \left( {2\sin^{2}A} \right)\]
Again, use the trigonometric ratio \[\csc x = \dfrac{1}{{\sin x }}\].
\[\csc A - 2\cot 2A \cos A = \dfrac{1}{{\sin A}} \left( {2\sin^{2}A} \right)\]
\[ \Rightarrow \csc A - 2\cot 2A \cos A = 2\sin A\]

Hence the correct option is A.

Note: Students often get confused about the trigonometric ratios.
Following are the basic trigonometric ratios:
\[\tan x = \dfrac{{\sin x }}{{\cos x }} = \dfrac{1}{{\cot x }}\]
\[\csc x = \dfrac{1}{{\sin x }}\]
\[\sec x = \dfrac{1}{{\cos x }}\]
\[\cot x = \dfrac{{\cos x}}{{\sin x }} = \dfrac{1}{{\tan x }}\]