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Question: How do you use the sum and difference formula to simplify \(\dfrac{1-\tan 40\tan 20}{\tan 40+\tan 20...

How do you use the sum and difference formula to simplify 1−tan⁡40tan⁡20tan⁡40+tan⁡20\dfrac{1-\tan 40\tan 20}{\tan 40+\tan 20} ?

Explanation

Solution

Here in this question, we will use the sum and difference formula of sine and cosine and derive the formula of tangent in order to solve such expressions. Sum and difference identities are:
⇒\Rightarrow sin( x + y) = sin(x)cos(y) + cos(x)sin(y)
⇒\Rightarrow cos(x + y) = cos(x)cos(y) – sin(x)sin(y)
⇒\Rightarrow tan(x + y) = tan⁡x+tan⁡y1−tan⁡xtan⁡y\dfrac{\tan x+\tan y}{1-\tan x\tan y}
⇒\Rightarrow sin(x - y) = sin(x)cos(y) – cos(x)sin(y)
⇒\Rightarrow cos(x - y) = cos(x)cos(y) + sin(x)sin(y)
⇒\Rightarrow tan(x - y) = tan⁡x−tan⁡y1+tan⁡xtan⁡y\dfrac{\tan x-\tan y}{1+\tan x\tan y}

Complete step by step answer:
Now, let’s solve the question.
The basic sum and difference formula for sin and cos are:
⇒\Rightarrow sin(x + y) = sin(x)cos(y) + cos(x)sin(y)
⇒\Rightarrow cos(x + y) = cos(x)cos(y) – sin(x)sin(y)
⇒\Rightarrow sin(x - y) = sin(x)cos(y) – cos(x)sin(y)
⇒\Rightarrow cos(x - y) = cos(x)cos(y) + sin(x)sin(y)
Now, we have to derive sum and difference formulae for tan from sin and cos above:
⇒\Rightarrow tan(x + y) = sin⁡(x+y)cos⁡(x+y)\dfrac{\sin (x+y)}{\cos (x+y)}
Now, place the formulae of sin(x + y) and cos(x + y). We get:
⇒\Rightarrow tan(x + y) = sin⁡xcos⁡y+cos⁡xsin⁡ycos⁡xcos⁡y−sin⁡xsin⁡y\dfrac{\sin x\cos y+\cos x\sin y}{\cos x\cos y-\sin x\sin y}
Next step is to divide the numerator and denominator of the left hand side with cos(x)cos(y). We will get:
⇒\Rightarrow tan(x + y) = sin⁡xcos⁡y+cos⁡xsin⁡ycos⁡xcos⁡ycos⁡xcos⁡y−sin⁡xsin⁡ycos⁡xcos⁡y\dfrac{\dfrac{\sin x\cos y+\cos x\sin y}{\cos x\cos y}}{\dfrac{\cos x\cos y-\sin x\sin y}{\cos x\cos y}}
Next, we have to reduce the like terms. We will get:
⇒\Rightarrow tan(x + y) = sin⁡xcos⁡x+sin⁡ycos⁡y1−sin⁡xsin⁡ycos⁡xcos⁡y\dfrac{\dfrac{\sin x}{\cos x}+\dfrac{\sin y}{\cos y}}{1-\dfrac{\sin x\sin y}{\cos x\cos y}}
As we know that tan⁡θ=sin⁡θcos⁡θ\tan \theta =\dfrac{\sin \theta }{\cos \theta }. So we will now substitute the value of tan in equation above:
⇒\Rightarrow tan(x + y) = tan⁡x+tan⁡y1−tan⁡xtan⁡y\dfrac{\tan x+\tan y}{1-\tan x\tan y}
Similarly, the derived formula for tan(x - y) will be:
⇒\Rightarrow tan(x - y) = tan⁡x−tan⁡y1+tan⁡xtan⁡y\dfrac{\tan x-\tan y}{1+\tan x\tan y}
Now, we need to write the formula for cot in terms of tan then the formula for cot(x + y) will be:
⇒\Rightarrow cot(x + y) = 1−tan⁡xtan⁡ytan⁡x+tan⁡y\dfrac{1-\tan x\tan y}{\tan x+\tan y}
And we are given in the question that:
⇒1−tan⁡40tan⁡20tan⁡40+tan⁡20\Rightarrow \dfrac{1-\tan 40\tan 20}{\tan 40+\tan 20}
By using the formula of cot(x + y), we will get:
⇒1−tan⁡40tan⁡20tan⁡40+tan⁡20\Rightarrow \dfrac{1-\tan 40\tan 20}{\tan 40+\tan 20} = cot(40 +20)
⇒\Rightarrow cot(60)
So, this is our final answer.

Note:
Students should remember all the sum and difference formulae for sin and cos and also know how to derive the formula for tan and cot. There is no need to find the value of cot(60) because we just have to simplify the expression. If you want to take out the value of cot(60) in the end, it’s completely fine. The marks will not be deducted.