Question

Prove the following:
$\cos \left( {\dfrac{{3\pi }}{2} + x} \right)\cos \left( {2\pi + x} \right)\left[ {\cot \left( {\dfrac{{3\pi }}{2} - x} \right) + \cot \left( {2\pi + x} \right)} \right] = 1$

Answer 1

We can take each of the terms in the LHS and simplify them. We can simplify the terms using the equations $\cos \left( {A + B} \right) = \cos \left( A \right)\cos \left( B \right) - \sin \left( A \right)\sin \left( B \right)$and $\cot \left( {A - B} \right) = \dfrac{{\cot A\cot B + 1}}{{\cot B - \cot A}}$. We can say that the equation is true when $LHS = RHS$

Complete step-by-step answer:
We need to prove$\cos \left( {\dfrac{{3\pi }}{2} + x} \right)\cos \left( {2\pi + x} \right)\left[ {\cot \left( {\dfrac{{3\pi }}{2} - x} \right) + \cot \left( {2\pi + x} \right)} \right] = 1$ .
We can simplify the terms in the LHS of the equation we need to prove.
We can take the 1st term, $\cos \left( {\dfrac{{3\pi }}{2} + x} \right)$
We can use the trigonometric identity, $\cos \left( {A + B} \right) = \cos \left( A \right)\cos \left( B \right) - \sin \left( A \right)\sin \left( B \right)$.
We can substitute the values,
$\Rightarrow \cos \left( {\dfrac{{3\pi }}{2} + x} \right) = \cos \left( {\dfrac{{3\pi }}{2}} \right)\cos \left( x \right) - \sin \left( {\dfrac{{3\pi }}{2}} \right)\sin \left( x \right)$
We know that $\cos \left( {\dfrac{{3\pi }}{2}} \right) = 0$and $\sin \left( {\dfrac{{3\pi }}{2}} \right) = - 1$. On substituting,
$\Rightarrow \cos \left( {\dfrac{{3\pi }}{2} + x} \right) = 0\cos \left( x \right) - \left( { - 1} \right)\sin \left( x \right)$
On simplification,
$\Rightarrow \cos \left( {\dfrac{{3\pi }}{2} + x} \right) = \sin \left( x \right)$ … (1)
We can take the 2nd term, $\cos \left( {2\pi + x} \right)$
We know that trigonometric functions are periodic and repeat the same values after the interval of $2\pi$.
$\cos \left( {2\pi + x} \right) = \cos \left( x \right)$ .. (2)
Now we can take the 3rd term, $\cot \left( {\dfrac{{3\pi }}{2} - x} \right)$,
We use the identity $\cot \left( {A - B} \right) = \dfrac{{\cot A\cot B + 1}}{{\cot B - \cot A}}$
We can substitute the values
$\Rightarrow \cot \left( {\dfrac{{3\pi }}{2} - x} \right) = \dfrac{{\cot \dfrac{{3\pi }}{2}\cot x + 1}}{{\cot x - \cot \dfrac{{3\pi }}{2}}}$
We know that $\cot \left( {\dfrac{{3\pi }}{2}} \right) = 0$. So, we get,
$\Rightarrow \cot \left( {\dfrac{{3\pi }}{2} - x} \right) = \dfrac{{0 \times \cot x + 1}}{{\cot x - 0}}$
On simplification,
$\Rightarrow \cot \left( {\dfrac{{3\pi }}{2} - x} \right) = \dfrac{1}{{\cot x}}$
We know that $\cot x = \dfrac{{\cos x}}{{\sin x}}$.
$\Rightarrow \cot \left( {\dfrac{{3\pi }}{2} - x} \right) = \dfrac{{\sin x}}{{\cos x}}$ … (3)
We can take the term, $\cot \left( {2\pi + x} \right)$
We know that trigonometric functions are periodic and repeat the same values after the interval of $2\pi$.
$\Rightarrow \cot \left( {2\pi + x} \right) = \cot \left( x \right)$
We know that $\cot x = \dfrac{{\cos x}}{{\sin x}}$.
$\Rightarrow \cot \left( {2\pi + x} \right) = \dfrac{{\cos x}}{{\sin x}}$… (4)
Now we can take the LHS,
$LHS = \cos \left( {\dfrac{{3\pi }}{2} + x} \right)\cos \left( {2\pi + x} \right)\left[ {\cot \left( {\dfrac{{3\pi }}{2} - x} \right) + \cot \left( {2\pi + x} \right)} \right]$
We can substitute equation (1), (2), (3) and (4) in the LHS,
$\Rightarrow LHS = \sin \left( x \right)\cos \left( x \right)\left[ {\dfrac{{\sin x}}{{\cos x}} + \dfrac{{\cos x}}{{\sin x}}} \right]$
After multiplication and simplification, we get,
$\Rightarrow LHS = \dfrac{{{{\sin }^2}x\cos x}}{{\cos x}} + \dfrac{{{{\cos }^2}x\sin x}}{{\sin x}} = {\sin ^2}x + {\cos ^2}x$
We know that, ${\sin ^2}x + {\cos ^2}x = 1$.
$\Rightarrow LHS = 1$.
RHS is also equal to 1. So, we can write,
$LHS = RHS$.
Hence the equation is proved.

Note: We must be familiar with the following trigonometric identities used in this problem.
1. $\cos \left( {A \pm B} \right) = \cos \left( A \right)\cos \left( B \right) \mp \sin \left( A \right)\sin \left( B \right)$
2.$\cot \left( {A \pm B} \right) = \dfrac{{\cot A\cot B \mp 1}}{{\cot B \pm \cot A}}$
3.${\sin ^2}x + {\cos ^2}x = 1$
4.$\cos \left( {2\pi + x} \right) = \cos \left( x \right)$