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# Prove: $\dfrac{{\cos 19^\circ - \sin 19^\circ }}{{\cos 19^\circ + \sin 19^\circ }} = \cot 64^\circ$

Last updated date: 18th Jun 2024
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Hint:
Here we need to prove the given trigonometric equation. For that, we will first consider the left hand side trigonometric expression and we will simplify the numerator and denominator using the basic trigonometric identities. After further simplification, we will get the expression equal to the right hand side expression.

Complete step by step solution:
First, we will consider the left hand side expression.
Now, we will divide the numerator and denominator by the term $\cos 19^\circ$.
$\Rightarrow \dfrac{{\cos 19^\circ - \sin 19^\circ }}{{\cos 19^\circ + \sin 19^\circ }} = \dfrac{{\dfrac{{\cos 19^\circ - \sin 19^\circ }}{{\cos 19^\circ }}}}{{\dfrac{{\cos 19^\circ + \sin 19^\circ }}{{\cos 19^\circ }}}}$
On further simplifying the numerator and denominator, we get
$\begin{array}{l} \Rightarrow \dfrac{{\cos 19^\circ - \sin 19^\circ }}{{\cos 19^\circ + \sin 19^\circ }} = \dfrac{{\dfrac{{\cos 19^\circ }}{{\cos 19^\circ }} - \dfrac{{\sin 19^\circ }}{{\cos 19^\circ }}}}{{\dfrac{{\cos 19^\circ }}{{\cos 19^\circ }} + \dfrac{{\sin 19^\circ }}{{\cos 19^\circ }}}}\\ \Rightarrow \dfrac{{\cos 19^\circ - \sin 19^\circ }}{{\cos 19^\circ + \sin 19^\circ }} = \dfrac{{1 - \dfrac{{\sin 19^\circ }}{{\cos 19^\circ }}}}{{1 + \dfrac{{\sin 19^\circ }}{{\cos 19^\circ }}}}\end{array}$
On applying the trigonometric identity $\dfrac{{\sin \theta }}{{\cos \theta }} = \tan \theta$ in the above equation, we get
$\Rightarrow \dfrac{{\cos 19^\circ - \sin 19^\circ }}{{\cos 19^\circ + \sin 19^\circ }} = \dfrac{{1 - \tan 19^\circ }}{{1 + \tan 19^\circ }}$
We know that $\tan 45^\circ = 1$ . Substituting this value in the equation, we get
$\Rightarrow \dfrac{{\cos 19^\circ - \sin 19^\circ }}{{\cos 19^\circ + \sin 19^\circ }} = \dfrac{{\tan 45^\circ - \tan 19^\circ }}{{1 + \tan 45^\circ \cdot \tan 19^\circ }}$
Using the trigonometric identity $\tan \left( {A - B} \right) = \dfrac{{\tan A - \tan B}}{{1 + \tan A \cdot \tan B}}$, we get
$\Rightarrow \dfrac{{\cos 19^\circ - \sin 19^\circ }}{{\cos 19^\circ + \sin 19^\circ }} = \tan \left( {45^\circ - 19^\circ } \right) = \tan 26^\circ$………………..$\left( 1 \right)$
We know from periodic identities that $\cot \left( {90^\circ - \theta } \right) = \tan \theta$.
Using this trigonometric identity, we can write
$\tan 26^\circ = \cot \left( {90^\circ - 26^\circ } \right)$
Now substituting value of $\tan 26^\circ$ in equation $\left( 1 \right)$, we get
$\begin{array}{l} \Rightarrow \dfrac{{\cos 19^\circ - \sin 19^\circ }}{{\cos 19^\circ + \sin 19^\circ }} = \cot \left( {90^\circ - 26^\circ } \right)\\ \Rightarrow \dfrac{{\cos 19^\circ - \sin 19^\circ }}{{\cos 19^\circ + \sin 19^\circ }} = \cot \left( {64^\circ } \right)\end{array}$
Hence, we have proved that $\dfrac{{\cos 19^\circ - \sin 19^\circ }}{{\cos 19^\circ + \sin 19^\circ }} = \cot 64^\circ$.

Note:
To solve this question, we need to keep in mind some basic trigonometric identities and formulae. Trigonometric identities are defined as the equalities which involve trigonometric functions. They are always true for every value of the occurring variables for which both sides of the equality are defined. It is important for us to remember that all the trigonometric identities are periodic in nature. They repeat their values after a certain interval. To prove any expression, we always consider the complex side of the expression and make it equal to the simple expression after solving it because this is the easiest way to prove any expression.