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Evaluate $\dfrac{1}{\sin{10^\circ}}-\dfrac{\sqrt{3}}{\cos{10^\circ}}$

The sine of angle ten degrees and cosine of angle ten degrees are involved in forming a trigonometric expression and we have to evaluate this trigonometric expression by simplification.


Simplify the Trigonometric expression

The two terms that contain trigonometric functions are connected by a minus sign and each term in the expression is in fraction form. Hence, the trigonometric expression can be simplified by the subtraction of fractions.

$=\,\,\,$ $\dfrac{1 \times \cos{(10^\circ)}-\sqrt{3} \times \sin{(10^\circ)}}{\sin{(10^\circ)} \times \cos{(10^\circ)}}$

Simplify the Expression in numerator

The rational expression contains trigonometric expressions in both numerator and denominator. So, we have to concentrate on simplifying each trigonometric expression. Now, let’s focus on simplifying the trigonometric expression in the numerator of the function.

$\dfrac{1 \times \cos{(10^\circ)}-\sqrt{3} \times \sin{(10^\circ)}}{\sin{(10^\circ)} \times \cos{(10^\circ)}}$

The trigonometric expression is formed by the subtraction and each term is in product form. It is a good idea to express it in sine angle difference form. Let’s check the possibility.

$=\,\,\,$ $\dfrac{1 \times \Big(1 \times \cos{(10^\circ)}-\sqrt{3} \times \sin{(10^\circ)}\Big)}{\sin{(10^\circ)} \times \cos{(10^\circ)}}$

$=\,\,\,$ $\dfrac{\dfrac{2}{2} \times \Big(1 \times \cos{(10^\circ)}-\sqrt{3} \times \sin{(10^\circ)}\Big)}{\sin{(10^\circ)} \times \cos{(10^\circ)}}$

$=\,\,\,$ $\dfrac{2 \times \Bigg(\dfrac{1 \times \cos{(10^\circ)}-\sqrt{3} \times \sin{(10^\circ)}}{2}\Bigg)}{\sin{(10^\circ)} \times \cos{(10^\circ)}}$

$=\,\,\,$ $\dfrac{2 \times \Bigg(\dfrac{1 \times \cos{(10^\circ)}}{2}-\dfrac{\sqrt{3} \times \sin{(10^\circ)}}{2}\Bigg)}{\sin{(10^\circ)} \times \cos{(10^\circ)}}$

$=\,\,\,$ $\dfrac{2 \times \Bigg(\dfrac{1}{2} \times \cos{(10^\circ)}-\dfrac{\sqrt{3}}{2} \times \sin{(10^\circ)}\Bigg)}{\sin{(10^\circ)} \times \cos{(10^\circ)}}$

According to trigonometry, sin of 30 degrees is quotient of one by two and cos of 30 degrees is quotient of square root of three by two.

$=\,\,\,$ $\dfrac{2 \times \Big(\sin{(30^\circ)} \times \cos{(10^\circ)}-\cos{(30^\circ)} \times \sin{(10^\circ)}\Big)}{\sin{(10^\circ)} \times \cos{(10^\circ)}}$

The trigonometric expression in the second factor of the numerator represents the expansion of sine angle difference formula. Therefore, that expression can be simplified as sin of difference of angles.

$=\,\,\,$ $\dfrac{2 \times \sin{(30^\circ-10^\circ)}}{\sin{(10^\circ)} \times \cos{(10^\circ)}}$

$=\,\,\,$ $\dfrac{2 \times \sin{(20^\circ)}}{\sin{(10^\circ)} \times \cos{(10^\circ)}}$

Simplify the Expression in denominator

Look at the expression in the denominator. The expression is the product of sine and cosine of same angle. When we see the product like this, you must remember the sine double angle identity but the denominator should consist of number two as a factor. So we have to include a number two as a factor in the expression acceptably and it allows us to simply the expression by the sine double angle identity.

$=\,\,\,$ $\dfrac{2 \times 2 \times \sin{(20^\circ)}}{2 \times \sin{(10^\circ)} \times \cos{(10^\circ)}}$

Now, we can use the sine double angle identity for simplifying the trigonometric expression in the denominator of the rational function.

$=\,\,\,$ $\dfrac{4 \times \sin{(20^\circ)}}{\sin{(2 \times 10^\circ)}}$

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

$=\,\,\, \require{cancel}$ $\dfrac{4 \times \cancel{\sin{(20^\circ)}}}{\cancel{\sin{(20^\circ)}}}$

$=\,\,\, 4$

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