Question

If $\sin {20^{\circ}} = p$ , then find the value of $\left( {\dfrac{{\sin {{380}^{\circ}} - \sin {{340}^{\circ}}}}{{\cos {{380}^{\circ}} + \cos {{340}^{\circ}}}}} \right)$ .
$\left( a \right){\text{ }}\sqrt {1 - {p^2}} $
$\left( b \right){\text{ }}\sqrt {\dfrac{{1 - {p^2}}}{p}} $
\[\left( c \right){\text{ }}\dfrac{p}{{\sqrt {1 - {p^2}} }}\]
\[\left( d \right){\text{ None of these}}\]

Answer 1

Hint:
For solving this type of question we just need to change the degree into the sums or difference of a number which can be used for the known degree. And by using the identity we can easily solve it. And for finding the $\cos \theta $ , we will use the formula ${\sin ^2}\theta + {\cos ^2}\theta = 1$ and in this way, we will solve it.

Formula used:
The trigonometric identity used are:
1) $sin\left( {{{360}^{\circ}} + \theta } \right) = \sin \theta $
2) $sin\left( {{{360}^{\circ}} - \theta } \right) = - \sin \theta $
3) $\cos \left( {{{360}^{\circ}} + \theta } \right) = \cos \theta $
4) $\cos \left( {{{360}^{\circ}} - \theta } \right) = \cos \theta $
Here, $\theta $ will be the angle.

Complete step by step solution:
So we have the trigonometric equation given as $ \Rightarrow \left( {\dfrac{{\sin {{380}^{\circ}} - \sin {{340}^{\circ}}}}{{\cos {{380}^{\circ}} + \cos {{340}^{\circ}}}}} \right)$
Now on changing the degree in such a way that we can use the identity, therefore we get
$ \Rightarrow \dfrac{{\sin \left( {{{360}^{\circ}} + {{20}^{\circ}}} \right) - \sin \left( {{{360}^{\circ}} - {{20}^{\circ}}} \right)}}{{\cos \left( {{{360}^{\circ}} + {{20}^{\circ}}} \right) + \cos \left( {{{360}^{\circ}} - {{20}^{\circ}}} \right)}}$
Now on applying the identity, we get
\[ \Rightarrow \dfrac{{\sin {{20}^{\circ}} - \left( { - \sin {{20}^{\circ}}} \right)}}{{\cos {{20}^{\circ}} + \cos {{20}^{\circ}}}}\]
So on solving the braces of the numerator we get
\[ \Rightarrow \dfrac{{\sin {{20}^{\circ}} + \sin {{20}^{\circ}}}}{{\cos {{20}^{\circ}} + \cos {{20}^{\circ}}}}\]
Now on solving the addition of numerator and denominator part of the above, we get
\[ \Rightarrow \dfrac{{2\sin {{20}^{\circ}}}}{{2\cos {{20}^{\circ}}}}\]
And since the same term will cancel each other, therefore,
\[ \Rightarrow \dfrac{{\sin {{20}^{\circ}}}}{{\cos {{20}^{\circ}}}}\] , and we will name it equation $1$
As the value of $\sin {20^{\circ}} = p$ is already given, so we have to first find the value of $\cos {20^{\circ}}$ and for this we know that ${\sin ^2}{20^{\circ}} + {\cos ^2}{20^{\circ}} = 1$
Therefore,
$ \Rightarrow {\cos ^2}{20^{\circ}} = 1 - {p^2}$
And on taking the square to the right side, we get
$ \Rightarrow \cos {20^{\circ}} = \sqrt {1 - {p^2}} $
Therefore, on solving the equation $1$ by substituting the required values, we get
$ \Rightarrow \dfrac{p}{{\sqrt {1 - {p^2}} }}$

Therefore, the option $\left( c \right)$ is correct.

Note:
So from this question we can see that for solving this type of question we just need to change the degree and by using some trigonometric identity we can easily solve this type of problem. Also, we have to memorize those formulas for solving the problem easily because without it we can’t solve it.