Question

Prove that - $\dfrac{{\sin 75^\circ + \sin 15^\circ }}{{\sin 75^\circ - \sin 15^\circ }} = \sqrt 3 $.

Answer 1

Hint: To solve this question we will use the addition property of sine functions that is,
$\sin A + \sin B = 2\sin \left( {\dfrac{{A + B}}{2}} \right)\cos \left( {\dfrac{{A - B}}{2}} \right)$
And also, the subtraction property of sine function, that is,
$\sin A - \sin B = 2\cos \left( {\dfrac{{A + B}}{2}} \right)\sin \left( {\dfrac{{A - B}}{2}} \right)$
Getting these values for the angles given in the question, then we will substitute the obtained values in the given condition. Then by operating accordingly we can get the answer to our problem. So, let us see how to solve this question.

Complete step-by-step answer:
To prove- $\dfrac{{\sin 75^\circ + \sin 15^\circ }}{{\sin 75^\circ - \sin 15^\circ }} = \sqrt 3 $.
LHS- $\dfrac{{\sin 75^\circ + \sin 15^\circ }}{{\sin 75^\circ - \sin 15^\circ }}$
Now, we know, the sine rule of addition, that is,
$\sin A + \sin B = 2\sin \left( {\dfrac{{A + B}}{2}} \right)\cos \left( {\dfrac{{A - B}}{2}} \right)$
Now, substituting $A = 75^\circ $ and $B = 15^\circ $, in the above formula, we get,
$\sin 75^\circ + \sin 15^\circ = 2\sin \left( {\dfrac{{75^\circ + 15^\circ }}{2}} \right)\cos \left( {\dfrac{{75^\circ - 15^\circ }}{2}} \right)$
$ \Rightarrow \sin 75^\circ + \sin 15^\circ = 2\sin \left( {\dfrac{{90^\circ }}{2}} \right)\cos \left( {\dfrac{{60^\circ }}{2}} \right)$
Further simplifying, we get,
$ \Rightarrow \sin 75^\circ + \sin 15^\circ = 2\sin \left( {45^\circ } \right)\cos \left( {30^\circ } \right) - - - \left( 1 \right)$
Again, we know, the sine rule of subtraction, that is,
$\sin A - \sin B = 2\cos \left( {\dfrac{{A + B}}{2}} \right)\sin \left( {\dfrac{{A - B}}{2}} \right)$
Now, substituting $A = 75^\circ $ and $B = 15^\circ $, in the above formula, we get,
$\sin 75^\circ - \sin 15^\circ = 2\cos \left( {\dfrac{{75^\circ + 15^\circ }}{2}} \right)\sin \left( {\dfrac{{75^\circ - 15^\circ }}{2}} \right)$
$ \Rightarrow \sin 75^\circ - \sin 15^\circ = 2\cos \left( {\dfrac{{90^\circ }}{2}} \right)\sin \left( {\dfrac{{60^\circ }}{2}} \right)$
Further simplifying, we get,
$ \Rightarrow \sin 75^\circ - \sin 15^\circ = 2\cos \left( {45^\circ } \right)\sin \left( {30^\circ } \right) - - - \left( 2 \right)$
Now, substituting the values obtained in $\left( 1 \right)$ and $\left( 2 \right)$ in $\dfrac{{\sin 75^\circ + \sin 15^\circ }}{{\sin 75^\circ - \sin 15^\circ }}$, we get,
$\dfrac{{\sin 75^\circ + \sin 15^\circ }}{{\sin 75^\circ - \sin 15^\circ }}$
\[ = \dfrac{{2\sin \left( {45^\circ } \right)\cos \left( {30^\circ } \right)}}{{2\cos \left( {45^\circ } \right)\sin \left( {30^\circ } \right)}}\]
\[ = \dfrac{{\sin \left( {45^\circ } \right)\cos \left( {30^\circ } \right)}}{{\cos \left( {45^\circ } \right)\sin \left( {30^\circ } \right)}}\]
Now, we know, $\dfrac{{\sin \theta }}{{\cos \theta }} = \tan \theta $ and $\dfrac{{\cos \theta }}{{\sin \theta }} = \cot \theta $.
Therefore, substituting these values in the above trigonometric sum, we get,
$ = \tan \left( {45^\circ } \right)\cot \left( {30^\circ } \right)$
Now, we know, $\tan 45^\circ = 1$ and $\cot 30^\circ = \sqrt 3 $.
Now, substituting these values in the above trigonometric sum, we get,
$ = 1.\sqrt 3 $
$ = \sqrt 3 $
Therefore, LHS=RHS.
That is, $\dfrac{{\sin 75^\circ + \sin 15^\circ }}{{\sin 75^\circ - \sin 15^\circ }} = \sqrt 3 $.
Hence, proved.

Note: We could have also used the values of sine and cosine function in the problem for the given values of the angles instead of turning them into tangent and cotangent functions. We would have the same answer in that way also. But we use the trigonometric formulas to simplify the expressions and find the final answer easily.