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Evaluate $\sqrt{3}\csc{20^\circ}-\sec{20^\circ}$

The secant of angle twenty degrees is subtracted from the square root of three times cosecant of angle twenty degrees to express a quantity in the form a trigonometric expression.


Let’s learn how to evaluate this trigonometric expression with understandable steps by simplification.

Converting the trigonometric expression

The angle $20^\circ$ is not a standard angle. So, it is not possible to remember the values of cosecant and secant of angle twenty degrees. In trigonometry, there are no identities in terms of secant and cosecant. Hence, it is essential to convert them into convenient form. According to the reciprocal trigonometric identities, the reciprocal of cosine can be written as secant and also the reciprocal of sine can be expressed as cosecant.

$=\,\,\,$ $\dfrac{\sqrt{3}}{\sin{20^\circ}}-\dfrac{1}{\cos{20^\circ}}$

Calculate the difference of the fractions

Now, the trigonometric expression represents the subtraction of two fractions. Hence, find their difference as per the difference rule of the fractions.

$=\,\,\,$ $\dfrac{\sqrt{3} \times \cos{20^\circ}-1 \times \sin{20^\circ}}{\sin{20^\circ} \times \cos{20^\circ}}$

Simplifying the trigonometric function

The rational function consists of two trigonometric expressions in both numerator and denominator. So, let us try to simplify the trigonometric expression in the numerator.

The cosine and sine with same angle are the factors in the terms of the expression and their coefficients are $\sqrt{3}$ and $1$. The numerical coefficients of them match with the numerator values of sine and cosine functions for the angle $60$ degrees. So, let us make some acceptable adjustments and it allows us simplify the trigonometric expression in the numerator as an expansion of the sine angle difference formula.

$=\,\,\,$ $\dfrac{1 \times \big(\sqrt{3} \times \cos{20^\circ}-1 \times \sin{20^\circ}\big)}{\sin{20^\circ}\cos{20^\circ}}$

$=\,\,\,$ $\dfrac{\dfrac{2}{2} \times \big(\sqrt{3} \times \cos{20^\circ}-1 \times \sin{20^\circ}\big)}{\sin{20^\circ}\cos{20^\circ}}$

$=\,\,\,$ $\dfrac{\dfrac{2 \times 1}{2} \times \big(\sqrt{3} \times \cos{20^\circ}-1 \times \sin{20^\circ}\big)}{\sin{20^\circ}\cos{20^\circ}}$

$=\,\,\,$ $\dfrac{2 \times \dfrac{1}{2} \times \big(\sqrt{3} \times \cos{20^\circ}-1 \times \sin{20^\circ}\big)}{\sin{20^\circ}\cos{20^\circ}}$

Now, multiply the factor $1$ by $2$ across the difference of the terms as per the distributive property of multiplication over multiplication.

$=\,\,\,$ $\dfrac{2 \times \bigg(\dfrac{1}{2} \times \sqrt{3} \times \cos{20^\circ}-\dfrac{1}{2} \times 1 \times \sin{20^\circ}\bigg)}{\sin{20^\circ}\cos{20^\circ}}$

$=\,\,\,$ $\dfrac{2 \times \bigg(\dfrac{1 \times \sqrt{3}}{2} \times \cos{20^\circ}-\dfrac{1 \times 1}{2} \times \sin{20^\circ}\bigg)}{\sin{20^\circ}\cos{20^\circ}}$

$=\,\,\,$ $\dfrac{2 \times \bigg(\dfrac{\sqrt{3}}{2} \times \cos{20^\circ}-\dfrac{1}{2} \times \sin{20^\circ}\bigg)}{\sin{20^\circ}\cos{20^\circ}}$

According to trigonometry, the sine of sixty degrees is equal to square root of three by two and cosine of sixty degrees is equal to one by two. So, the numbers in fraction form can be replaced by their respective trigonometric ratios.

$=\,\,\,$ $\dfrac{2 \times \big(\sin{60^\circ} \times \cos{20^\circ}-\cos{60^\circ} \times \sin{20^\circ}\big)}{\sin{20^\circ}\cos{20^\circ}}$

$=\,\,\,$ $\dfrac{2 \times \big(\sin{60^\circ}\cos{20^\circ}-\cos{60^\circ}\sin{20^\circ}\big)}{\sin{20^\circ}\cos{20^\circ}}$

The second factor in the trigonometric expression of the numerator represents the expansion of sine angle different identity exactly. So, it can be simplified as the sine of angle difference between the $60^\circ$ and $20^\circ$.

$=\,\,\,$ $\dfrac{2 \times \sin{(60^\circ-20^\circ)}}{\sin{20^\circ}\cos{20^\circ}}$

$=\,\,\,$ $2 \times \dfrac{\sin{40^\circ}}{\sin{20^\circ}\cos{20^\circ}}$

Now, let’s focus on simplifying the trigonometric expression in the denominator. The factors sine and cosine consist of same angle, their product represents expansion of sine double angle identity but it should have the number $2$ as a numerical coefficient. So, let’s try to make some adjustments to get the exact form of sine double angle identity.

$=\,\,\,$ $2 \times 1 \times \dfrac{\sin{40^\circ}}{\sin{20^\circ}\cos{20^\circ}}$

$=\,\,\,$ $2 \times \dfrac{2}{2} \times \dfrac{\sin{40^\circ}}{\sin{20^\circ}\cos{20^\circ}}$

$=\,\,\,$ $2 \times \dfrac{2 \times 1}{2} \times \dfrac{\sin{40^\circ}}{\sin{20^\circ}\cos{20^\circ}}$

$=\,\,\,$ $2 \times 2 \times \dfrac{1}{2} \times \dfrac{\sin{40^\circ}}{\sin{20^\circ}\cos{20^\circ}}$

$=\,\,\,$ $4 \times \dfrac{1}{2} \times \dfrac{\sin{40^\circ}}{\sin{20^\circ}\cos{20^\circ}}$

Now, multiply the fractions to get the denominator as an expansion of sine double angle rule.

$=\,\,\,$ $4 \times \dfrac{1 \times \sin{40^\circ}}{2 \times \sin{20^\circ}\cos{20^\circ}}$

$=\,\,\,$ $4 \times \dfrac{\sin{40^\circ}}{2\sin{20^\circ}\cos{20^\circ}}$

The trigonometric expression in the denominator is exactly the expansion of sine of double angle formula but the angle is twenty degrees.

$=\,\,\,$ $4 \times \dfrac{\sin{40^\circ}}{\sin{(2 \times 20^\circ)}}$

$=\,\,\,$ $4 \times \dfrac{\sin{40^\circ}}{\sin{40^\circ}}$

$=\,\,\,$ $4 \times \dfrac{\cancel{\sin{40^\circ}}}{\cancel{\sin{40^\circ}}}$

$=\,\,\,$ $4 \times 1$

$=\,\,\,$ $4$

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