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Question: How do you simplify \(\dfrac{{\sin {{80}^0} - \sin {{10}^0}}}{{\sin {{80}^0} + \sin {{10}^0}}}\)?...

How do you simplify sin⁡800−sin⁡100sin⁡800+sin⁡100\dfrac{{\sin {{80}^0} - \sin {{10}^0}}}{{\sin {{80}^0} + \sin {{10}^0}}}?

Explanation

Solution

We are given and trigonometric expression in the form of fraction. In numerator the difference of two Sine angles is given and its numerator is the sum of two Sine angles. To solve the expression we will apply the formula of Sin A-Sin B in the numerator and Sin A + Sin B in the denominator.
Formula used:
SinA−SinB=2Cos(A+B)2×SinA−B2Sin A- Sin B= \dfrac{2Cos(A+B)}{2} \times \dfrac{SinA-B}{2}
SinA+SinB=2Sin(A+B)2×Cos(A−B)2Sin A + Sin B= \dfrac{2Sin(A+B)}{2} \times \dfrac{Cos(A-B)}{2}
By using this formula we can solve the numerator and denominator and then we will substitute the value of a trigonometric angle from the table. And then we will use the trigonometric ratios i.e.
TanA=SinACosATan A = \dfrac{SinA}{CosA}

Complete step-by-step answer:
Step1: We are given a trigonometric expressionsin⁡800−sin⁡100sin⁡800+sin⁡100\dfrac{{\sin {{80}^0} - \sin {{10}^0}}}{{\sin {{80}^0} + \sin {{10}^0}}} to solve the numerator and denominator we will apply the formula of
SinA−SinB=2Cos(A+B)2×SinA−B2Sin A- Sin B= \dfrac{2Cos(A+B)}{2} \times \dfrac{SinA-B}{2} in numerator and for denominator we will use the formula SinA+SinB=2Sin(A+B)2×Cos(A−B)2Sin A + Sin B= \dfrac{2Sin(A+B)}{2} \times \dfrac{Cos(A-B)}{2}
Here Sin⁡A=Sin⁡80\operatorname{Sin} A = \operatorname{Sin} 80 and Sin⁡B=Sin⁡10\operatorname{Sin} B = \operatorname{Sin} 10.
On substituting the values in the formula we will get:
⇒2Cos⁡(80+102)Sin⁡(80−102)2Sin⁡(80+102)Cos⁡(80−102)\Rightarrow \dfrac{{2\operatorname{Cos} \left( {\dfrac{{80 + 10}}{2}} \right)\operatorname{Sin} \left( {\dfrac{{80 - 10}}{2}} \right)}}{{2\operatorname{Sin} \left( {\dfrac{{80 + 10}}{2}} \right)\operatorname{Cos} \left( {\dfrac{{80 - 10}}{2}} \right)}}
Step2: On further solving it we get:
⇒2Cos⁡45.Sin⁡352Sin⁡45.Cos⁡35\Rightarrow \dfrac{{2\operatorname{Cos} 45.\operatorname{Sin} 35}}{{2\operatorname{Sin} 45.\operatorname{Cos} 35}}
On cancelling 2 from numerator and denominator we will get:
⇒Cos⁡45.Sin⁡35Sin⁡45.Cos⁡35\Rightarrow \dfrac{{\operatorname{Cos} 45.\operatorname{Sin} 35}}{{\operatorname{Sin} 45.\operatorname{Cos} 35}}
On using the trigonometric ratios that
Tan A =Sin⁡ACos⁡A;CotA=Cos⁡ASin⁡ATan{\text{ }}A{\text{ }} = \dfrac{{\operatorname{Sin} A}}{{\operatorname{Cos} A}};CotA = \dfrac{{\operatorname{Cos} A}}{{\operatorname{Sin} A}}We will get:
⇒Cot45.Tan⁡35\Rightarrow Cot45.\operatorname{Tan} 35
Substituting the value of Cot45=1Cot45 = 1 from the table we will get:
Tan⁡35\operatorname{Tan} 35

Hence the answer is Tan⁡35\operatorname{Tan} 35

Note:
In this type of questions students mainly apply the concept of trigonometric ratios of complementary angles which is not possible except in some cases. The best way to solve such questions is to directly apply the formula of the trigonometric angle which is given. Whether the angle is sin, cos or tan. Use the formula and solve the expression. And after this you can also use the values of angles from Tables. By this procedure the question will not get wrong.
Commit to memory:
SinA−SinB=2Cos(A+B)2×SinA−B2Sin A- Sin B= \dfrac{2Cos(A+B)}{2} \times \dfrac{SinA-B}{2}
SinA+SinB=2Sin(A+B)2×Cos(A−B)2Sin A + Sin B= \dfrac{2Sin(A+B)}{2} \times \dfrac{Cos(A-B)}{2}