Evaluate $\dfrac{5\sin{(17^\circ)}}{\cos{(73^\circ)}}$ $+$ $\dfrac{2\cos{(31^\circ)}}{\sin{(59^\circ)}}$ $-$ $\dfrac{7\sin{(80^\circ)}}{\cos{(10^\circ)}}$

In this trigonometry problem, the trigonometric functions sine and cosine with different angles in ratio form formed a trigonometric expression as follows.

$\dfrac{5\sin{(17^\circ)}}{\cos{(73^\circ)}}$ $+$ $\dfrac{2\cos{(31^\circ)}}{\sin{(59^\circ)}}$ $-$ $\dfrac{7\sin{(80^\circ)}}{\cos{(10^\circ)}}$

In this trigonometric expression, $5$ times the quotient of sine of angle $17$ degrees by cosine of $73$ degrees is added to the $2$ times the quotient of cosine of angle $31$ degrees by sine of angle $59$ degrees. $7$ times the quotient of sine of angle $80$ degrees by cosine of angle $10$ degrees is subtracted from the sum of the terms.

Now, let us learn how to evaluate the given trigonometric expression.

Evaluate the Quotient in first term

Let us focus on the first term and try to evaluate the value of this term.

In the first term, the sine of angle $17$ degrees and cosine of angle $73$ degrees should be substituted to find their quotient but their values are not known to us.

$=\,\,\,$ $\dfrac{5 \times \sin{(17^\circ)}}{\cos{(73^\circ)}}$ $+$ $\dfrac{2\cos{(31^\circ)}}{\sin{(59^\circ)}}$ $-$ $\dfrac{7\sin{(80^\circ)}}{\cos{(10^\circ)}}$

$=\,\,\,$ $5 \times \dfrac{\sin{(17^\circ)}}{\cos{(73^\circ)}}$ $+$ $\dfrac{2\cos{(31^\circ)}}{\sin{(59^\circ)}}$ $-$ $\dfrac{7\sin{(80^\circ)}}{\cos{(10^\circ)}}$

The value of first term can be calculated without knowing them because the angles in sine and cosine functions in first term are complementary. Hence, express an angle in its complementary form.

$=\,\,\,$ $5 \times \dfrac{\sin{(17^\circ)}}{\cos{(90^\circ-17^\circ)}}$ $+$ $\dfrac{2\cos{(31^\circ)}}{\sin{(59^\circ)}}$ $-$ $\dfrac{7\sin{(80^\circ)}}{\cos{(10^\circ)}}$

As per the cofunction identity of cosine function, the cosine of angle can be written as the sine of the complementary angle.

$=\,\,\,$ $5 \times \dfrac{\sin{(17^\circ)}}{\sin{(17^\circ)}}$ $+$ $\dfrac{2\cos{(31^\circ)}}{\sin{(59^\circ)}}$ $-$ $\dfrac{7\sin{(80^\circ)}}{\cos{(10^\circ)}}$

$=\,\,\,$ $5 \times \dfrac{\cancel{\sin{(17^\circ)}}}{\cancel{\sin{(17^\circ)}}}$ $+$ $\dfrac{2\cos{(31^\circ)}}{\sin{(59^\circ)}}$ $-$ $\dfrac{7\sin{(80^\circ)}}{\cos{(10^\circ)}}$

$=\,\,\,$ $5 \times 1$ $+$ $\dfrac{2\cos{(31^\circ)}}{\sin{(59^\circ)}}$ $-$ $\dfrac{7\sin{(80^\circ)}}{\cos{(10^\circ)}}$

$=\,\,\,$ $5$ $+$ $\dfrac{2\cos{(31^\circ)}}{\sin{(59^\circ)}}$ $-$ $\dfrac{7\sin{(80^\circ)}}{\cos{(10^\circ)}}$

Evaluate the Quotient in second term

It is time to find the value of second term. The cosine of angle $31$ degrees and sine of angle $59$ degrees have to substitute in the second term but their values are not known to us.

$=\,\,\,$ $5$ $+$ $\dfrac{2 \times \cos{(31^\circ)}}{\sin{(59^\circ)}}$ $-$ $\dfrac{7\sin{(80^\circ)}}{\cos{(10^\circ)}}$

$=\,\,\,$ $5$ $+$ $2 \times \dfrac{\cos{(31^\circ)}}{\sin{(59^\circ)}}$ $-$ $\dfrac{7\sin{(80^\circ)}}{\cos{(10^\circ)}}$

However, the value of quotient in the second term can be evaluated without knowing their values because the angles in sine and cosine functions are complementary angles. Therefore, write an angle in its complementary angle form.

$=\,\,\,$ $5$ $+$ $2 \times \dfrac{\cos{(90^\circ-59^\circ)}}{\sin{(59^\circ)}}$ $-$ $\dfrac{7\sin{(80^\circ)}}{\cos{(10^\circ)}}$

Once again, use the cofunction identity of the cosine function to calculate the value of second term.

$=\,\,\,$ $5$ $+$ $2 \times \dfrac{\sin{(59^\circ)}}{\sin{(59^\circ)}}$ $-$ $\dfrac{7\sin{(80^\circ)}}{\cos{(10^\circ)}}$

$=\,\,\,$ $5$ $+$ $2 \times \dfrac{\cancel{\sin{(59^\circ)}}}{\cancel{\sin{(59^\circ)}}}$ $-$ $\dfrac{7\sin{(80^\circ)}}{\cos{(10^\circ)}}$

$=\,\,\,$ $5$ $+$ $2 \times 1$ $-$ $\dfrac{7\sin{(80^\circ)}}{\cos{(10^\circ)}}$

$=\,\,\,$ $5$ $+$ $2$ $-$ $\dfrac{7\sin{(80^\circ)}}{\cos{(10^\circ)}}$

$=\,\,\,$ $7$ $-$ $\dfrac{7\sin{(80^\circ)}}{\cos{(10^\circ)}}$

Evaluate the Quotient in third term

Finally, it is time to find the value of third term. The sine of angle $80$ degrees and cosine of angle $10$ degrees should be substituted to find the value of third term.

$=\,\,\,$ $7$ $-$ $\dfrac{7 \times \sin{(80^\circ)}}{\cos{(10^\circ)}}$

$=\,\,\,$ $7$ $-$ $7 \times \dfrac{\sin{(80^\circ)}}{\cos{(10^\circ)}}$

The value of quotient in the third term can be calculated without knowing the values of the trigonometric functions because the angles in the cosine and sine functions are complementary. Hence, write any angle as the complementary of another angle.

$=\,\,\,$ $7$ $-$ $7 \times \dfrac{\sin{(90^\circ-10^\circ)}}{\cos{(10^\circ)}}$

Use the cofunction identity of sine function to convert the sine function in cosine function.

$=\,\,\,$ $7$ $-$ $7 \times \dfrac{\cos{(10^\circ)}}{\cos{(10^\circ)}}$

$=\,\,\,$ $7$ $-$ $7 \times \dfrac{\cancel{\cos{(10^\circ)}}}{\cancel{\cos{(10^\circ)}}}$

$=\,\,\,$ $7$ $-$ $7 \times 1$

$=\,\,\,$ $7-7$

$=\,\,\,$ $0$