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Question: If we are given a trigonometric expression as \(\tan \left( \alpha \right)=K\cot \left( \beta \right...

If we are given a trigonometric expression as tan⁡(α)=Kcot⁡(β)\tan \left( \alpha \right)=K\cot \left( \beta \right), then the value of cos⁡(α−β)cos⁡(α+β)\dfrac{\cos \left( \alpha -\beta \right)}{\cos \left( \alpha +\beta \right)} is equals to
(a) 1+K1−K\dfrac{1+K}{1-K}
(b) 1−K1+K\dfrac{1-K}{1+K}
(c) K+1K−1\dfrac{K+1}{K-1}
(d) K−1K+1\dfrac{K-1}{K+1}

Explanation

Solution

Hint: We will apply the formulas cos⁡(α−β)=cos⁡(α)cos⁡(β)+sin⁡(α)sin⁡(β)\cos \left( \alpha -\beta \right)=\cos \left( \alpha \right)\cos \left( \beta \right)+\sin \left( \alpha \right)\sin \left( \beta \right) and cos⁡(α+β)=cos⁡(α)cos⁡(β)−sin⁡(α)sin⁡(β)\cos \left( \alpha +\beta \right)=\cos \left( \alpha \right)\cos \left( \beta \right)-\sin \left( \alpha \right)\sin \left( \beta \right) to solve the question. We will also take the help of trigonometric conversions like tan⁡(α)=sin⁡(α)cos⁡(α)\tan \left( \alpha \right)=\dfrac{\sin \left( \alpha \right)}{\cos \left( \alpha \right)} and cot⁡(β)=cos⁡(β)sin⁡(β)\cot \left( \beta \right)=\dfrac{\cos \left( \beta \right)}{\sin \left( \beta \right)} to convert the given trigonometric expression in the question into simpler terms.

Complete step-by-step answer:

Now, we will consider the given trigonometric expression tan⁡(α)=Kcot⁡(β)\tan \left( \alpha \right)=K\cot \left( \beta \right) and take cotangent terms to the right side of the equation. Thus, we will get K=tan⁡(α)cot⁡(β)K=\dfrac{\tan \left( \alpha \right)}{\cot \left( \beta \right)}.
Now, we will start by converting it into simpler terms. We will substitute tan⁡(α)=sin⁡(α)cos⁡(α)\tan \left( \alpha \right)=\dfrac{\sin \left( \alpha \right)}{\cos \left( \alpha \right)} and cot⁡(β)=cos⁡(β)sin⁡(β)\cot \left( \beta \right)=\dfrac{\cos \left( \beta \right)}{\sin \left( \beta \right)} in the equation we will get K=sin⁡(α)cos⁡(α)cos⁡(β)sin⁡(β)K=\dfrac{\dfrac{\sin \left( \alpha \right)}{\cos \left( \alpha \right)}}{\dfrac{\cos \left( \beta \right)}{\sin \left( \beta \right)}}.
Now, we will apply the property which is given by abcd=adcb\dfrac{\dfrac{a}{b}}{\dfrac{c}{d}}=\dfrac{ad}{cb} in the equation and we will get K=sin⁡(α)sin⁡(β)cos⁡(α)cos⁡(β)K=\dfrac{\sin \left( \alpha \right)\sin \left( \beta \right)}{\cos \left( \alpha \right)\cos \left( \beta \right)}.
Now, we will come to the expression cos⁡(α−β)cos⁡(α+β)\dfrac{\cos \left( \alpha -\beta \right)}{\cos \left( \alpha +\beta \right)}. By the formula cos⁡(α−β)=cos⁡(α)cos⁡(β)+sin⁡(α)sin⁡(β)\cos \left( \alpha -\beta \right)=\cos \left( \alpha \right)\cos \left( \beta \right)+\sin \left( \alpha \right)\sin \left( \beta \right) and cos⁡(α+β)=cos⁡(α)cos⁡(β)−sin⁡(α)sin⁡(β)\cos \left( \alpha +\beta \right)=\cos \left( \alpha \right)\cos \left( \beta \right)-\sin \left( \alpha \right)\sin \left( \beta \right) we will get cos⁡(α−β)cos⁡(α+β)=cos⁡(α)cos⁡(β)+sin⁡(α)sin⁡(β)cos⁡(α)cos⁡(β)−sin⁡(α)sin⁡(β)\dfrac{\cos \left( \alpha -\beta \right)}{\cos \left( \alpha +\beta \right)}=\dfrac{\cos \left( \alpha \right)\cos \left( \beta \right)+\sin \left( \alpha \right)\sin \left( \beta \right)}{\cos \left( \alpha \right)\cos \left( \beta \right)-\sin \left( \alpha \right)\sin \left( \beta \right)}.
Now we will divide the numerator and denominator by cos⁡(α)cos⁡(β)\cos \left( \alpha \right)\cos \left( \beta \right) therefore, we get cos⁡(α−β)cos⁡(α+β)=cos⁡(α)cos⁡(β)+sin⁡(α)sin⁡(β)cos⁡(α)cos⁡(β)cos⁡(α)cos⁡(β)−sin⁡(α)sin⁡(β)cos⁡(α)cos⁡(β) ⇒cos⁡(α−β)cos⁡(α+β)=cos⁡(α)cos⁡(β)cos⁡(α)cos⁡(β)+sin⁡(α)sin⁡(β)cos⁡(α)cos⁡(β)cos⁡(α)cos⁡(β)cos⁡(α)cos⁡(β)−sin⁡(α)sin⁡(β)cos⁡(α)cos⁡(β) ⇒cos⁡(α−β)cos⁡(α+β)=1+sin⁡(α)sin⁡(β)cos⁡(α)cos⁡(β)1−sin⁡(α)sin⁡(β)cos⁡(α)cos⁡(β) \begin{aligned} & \dfrac{\cos \left( \alpha -\beta \right)}{\cos \left( \alpha +\beta \right)}=\dfrac{\dfrac{\cos \left( \alpha \right)\cos \left( \beta \right)+\sin \left( \alpha \right)\sin \left( \beta \right)}{\cos \left( \alpha \right)\cos \left( \beta \right)}}{\dfrac{\cos \left( \alpha \right)\cos \left( \beta \right)-\sin \left( \alpha \right)\sin \left( \beta \right)}{\cos \left( \alpha \right)\cos \left( \beta \right)}} \\\ & \Rightarrow \dfrac{\cos \left( \alpha -\beta \right)}{\cos \left( \alpha +\beta \right)}=\dfrac{\dfrac{\cos \left( \alpha \right)\cos \left( \beta \right)}{\cos \left( \alpha \right)\cos \left( \beta \right)}+\dfrac{\sin \left( \alpha \right)\sin \left( \beta \right)}{\cos \left( \alpha \right)\cos \left( \beta \right)}}{\dfrac{\cos \left( \alpha \right)\cos \left( \beta \right)}{\cos \left( \alpha \right)\cos \left( \beta \right)}-\dfrac{\sin \left( \alpha \right)\sin \left( \beta \right)}{\cos \left( \alpha \right)\cos \left( \beta \right)}} \\\ & \Rightarrow \dfrac{\cos \left( \alpha -\beta \right)}{\cos \left( \alpha +\beta \right)}=\dfrac{1+\dfrac{\sin \left( \alpha \right)\sin \left( \beta \right)}{\cos \left( \alpha \right)\cos \left( \beta \right)}}{1-\dfrac{\sin \left( \alpha \right)\sin \left( \beta \right)}{\cos \left( \alpha \right)\cos \left( \beta \right)}} \\\ \end{aligned}
As we know that K=sin⁡(α)sin⁡(β)cos⁡(α)cos⁡(β)K=\dfrac{\sin \left( \alpha \right)\sin \left( \beta \right)}{\cos \left( \alpha \right)\cos \left( \beta \right)} therefore we get cos⁡(α−β)cos⁡(α+β)=1+K1−K\dfrac{\cos \left( \alpha -\beta \right)}{\cos \left( \alpha +\beta \right)}=\dfrac{1+K}{1-K}. Hence, the correct option is (a).

Note: The other way of solving this question is that we can put the value of K one by one in the options and lead it to the respected cosine formula. Whichever option is satisfied and results into cos⁡(α−β)cos⁡(α+β)\dfrac{\cos \left( \alpha -\beta \right)}{\cos \left( \alpha +\beta \right)} will be the correct answer.
For example we take the option (a) and substitute K=sin⁡(α)sin⁡(β)cos⁡(α)cos⁡(β)K=\dfrac{\sin \left( \alpha \right)\sin \left( \beta \right)}{\cos \left( \alpha \right)\cos \left( \beta \right)} in it will result into,
1+K1−K=1+sin⁡(α)sin⁡(β)cos⁡(α)cos⁡(β)1−sin⁡(α)sin⁡(β)cos⁡(α)cos⁡(β) ⇒1+K1−K=cos⁡(α)cos⁡(β)+sin⁡(α)sin⁡(β)cos⁡(α)cos⁡(β)cos⁡(α)cos⁡(β)−sin⁡(α)sin⁡(β)cos⁡(α)cos⁡(β) ⇒1+K1−K=cos⁡(α)cos⁡(β)+sin⁡(α)sin⁡(β)cos⁡(α)cos⁡(β)−sin⁡(α)sin⁡(β) \begin{aligned} & \dfrac{1+K}{1-K}=\dfrac{1+\dfrac{\sin \left( \alpha \right)\sin \left( \beta \right)}{\cos \left( \alpha \right)\cos \left( \beta \right)}}{1-\dfrac{\sin \left( \alpha \right)\sin \left( \beta \right)}{\cos \left( \alpha \right)\cos \left( \beta \right)}} \\\ & \Rightarrow \dfrac{1+K}{1-K}=\dfrac{\dfrac{\cos \left( \alpha \right)\cos \left( \beta \right)+\sin \left( \alpha \right)\sin \left( \beta \right)}{\cos \left( \alpha \right)\cos \left( \beta \right)}}{\dfrac{\cos \left( \alpha \right)\cos \left( \beta \right)-\sin \left( \alpha \right)\sin \left( \beta \right)}{\cos \left( \alpha \right)\cos \left( \beta \right)}} \\\ & \Rightarrow \dfrac{1+K}{1-K}=\dfrac{\cos \left( \alpha \right)\cos \left( \beta \right)+\sin \left( \alpha \right)\sin \left( \beta \right)}{\cos \left( \alpha \right)\cos \left( \beta \right)-\sin \left( \alpha \right)\sin \left( \beta \right)} \\\ \end{aligned}
By the formulas cos⁡(α−β)=cos⁡(α)cos⁡(β)+sin⁡(α)sin⁡(β)\cos \left( \alpha -\beta \right)=\cos \left( \alpha \right)\cos \left( \beta \right)+\sin \left( \alpha \right)\sin \left( \beta \right) and cos⁡(α+β)=cos⁡(α)cos⁡(β)−sin⁡(α)sin⁡(β)\cos \left( \alpha +\beta \right)=\cos \left( \alpha \right)\cos \left( \beta \right)-\sin \left( \alpha \right)\sin \left( \beta \right) we will get cos⁡(α−β)cos⁡(α+β)=1+K1−K\dfrac{\cos \left( \alpha -\beta \right)}{\cos \left( \alpha +\beta \right)}=\dfrac{1+K}{1-K} which is the required answer here. We can also divide the numerator and denominator of the equation cos⁡(α−β)cos⁡(α+β)=cos⁡(α)cos⁡(β)+sin⁡(α)sin⁡(β)cos⁡(α)cos⁡(β)−sin⁡(α)sin⁡(β)\dfrac{\cos \left( \alpha -\beta \right)}{\cos \left( \alpha +\beta \right)}=\dfrac{\cos \left( \alpha \right)\cos \left( \beta \right)+\sin \left( \alpha \right)\sin \left( \beta \right)}{\cos \left( \alpha \right)\cos \left( \beta \right)-\sin \left( \alpha \right)\sin \left( \beta \right)} by sin⁡(α)sin⁡(β)\sin \left( \alpha \right)\sin \left( \beta \right) also and we will get to the same result.