Question

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

Answer 1

Hint: We will apply the formulas $\cos \left( \alpha -\beta \right)=\cos \left( \alpha \right)\cos \left( \beta \right)+\sin \left( \alpha \right)\sin \left( \beta \right)$ and $\cos \left( \alpha +\beta \right)=\cos \left( \alpha \right)\cos \left( \beta \right)-\sin \left( \alpha \right)\sin \left( \beta \right)$ to solve the question. We will also take the help of trigonometric conversions like $\tan \left( \alpha \right)=\dfrac{\sin \left( \alpha \right)}{\cos \left( \alpha \right)}$ and $\cot \left( \beta \right)=\dfrac{\cos \left( \beta \right)}{\sin \left( \beta \right)}$ 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 $\tan \left( \alpha \right)=K\cot \left( \beta \right)$ and take cotangent terms to the right side of the equation. Thus, we will get $K=\dfrac{\tan \left( \alpha \right)}{\cot \left( \beta \right)}$.
Now, we will start by converting it into simpler terms. We will substitute $\tan \left( \alpha \right)=\dfrac{\sin \left( \alpha \right)}{\cos \left( \alpha \right)}$ and $\cot \left( \beta \right)=\dfrac{\cos \left( \beta \right)}{\sin \left( \beta \right)}$ in the equation we will get $K=\dfrac{\dfrac{\sin \left( \alpha \right)}{\cos \left( \alpha \right)}}{\dfrac{\cos \left( \beta \right)}{\sin \left( \beta \right)}}$.
Now, we will apply the property which is given by $\dfrac{\dfrac{a}{b}}{\dfrac{c}{d}}=\dfrac{ad}{cb}$ in the equation and we will get $K=\dfrac{\sin \left( \alpha \right)\sin \left( \beta \right)}{\cos \left( \alpha \right)\cos \left( \beta \right)}$.
Now, we will come to the expression $\dfrac{\cos \left( \alpha -\beta \right)}{\cos \left( \alpha +\beta \right)}$. By the formula $\cos \left( \alpha -\beta \right)=\cos \left( \alpha \right)\cos \left( \beta \right)+\sin \left( \alpha \right)\sin \left( \beta \right)$ and $\cos \left( \alpha +\beta \right)=\cos \left( \alpha \right)\cos \left( \beta \right)-\sin \left( \alpha \right)\sin \left( \beta \right)$ we will get $\dfrac{\cos \left( \alpha -\beta \right)}{\cos \left( \alpha +\beta \right)}=\dfrac{\cos \left( \alpha \right)\cos \left( \beta \right)+\sin \left( \alpha \right)\sin \left( \beta \right)}{\cos \left( \alpha \right)\cos \left( \beta \right)-\sin \left( \alpha \right)\sin \left( \beta \right)}$.
Now we will divide the numerator and denominator by $\cos \left( \alpha \right)\cos \left( \beta \right)$ therefore, we get $\begin{align}
  & \dfrac{\cos \left( \alpha -\beta \right)}{\cos \left( \alpha +\beta \right)}=\dfrac{\dfrac{\cos \left( \alpha \right)\cos \left( \beta \right)+\sin \left( \alpha \right)\sin \left( \beta \right)}{\cos \left( \alpha \right)\cos \left( \beta \right)}}{\dfrac{\cos \left( \alpha \right)\cos \left( \beta \right)-\sin \left( \alpha \right)\sin \left( \beta \right)}{\cos \left( \alpha \right)\cos \left( \beta \right)}} \\
 & \Rightarrow \dfrac{\cos \left( \alpha -\beta \right)}{\cos \left( \alpha +\beta \right)}=\dfrac{\dfrac{\cos \left( \alpha \right)\cos \left( \beta \right)}{\cos \left( \alpha \right)\cos \left( \beta \right)}+\dfrac{\sin \left( \alpha \right)\sin \left( \beta \right)}{\cos \left( \alpha \right)\cos \left( \beta \right)}}{\dfrac{\cos \left( \alpha \right)\cos \left( \beta \right)}{\cos \left( \alpha \right)\cos \left( \beta \right)}-\dfrac{\sin \left( \alpha \right)\sin \left( \beta \right)}{\cos \left( \alpha \right)\cos \left( \beta \right)}} \\
 & \Rightarrow \dfrac{\cos \left( \alpha -\beta \right)}{\cos \left( \alpha +\beta \right)}=\dfrac{1+\dfrac{\sin \left( \alpha \right)\sin \left( \beta \right)}{\cos \left( \alpha \right)\cos \left( \beta \right)}}{1-\dfrac{\sin \left( \alpha \right)\sin \left( \beta \right)}{\cos \left( \alpha \right)\cos \left( \beta \right)}} \\
\end{align}$
As we know that $K=\dfrac{\sin \left( \alpha \right)\sin \left( \beta \right)}{\cos \left( \alpha \right)\cos \left( \beta \right)}$ therefore we get $\dfrac{\cos \left( \alpha -\beta \right)}{\cos \left( \alpha +\beta \right)}=\dfrac{1+K}{1-K}$. 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 $\dfrac{\cos \left( \alpha -\beta \right)}{\cos \left( \alpha +\beta \right)}$ will be the correct answer.
For example we take the option (a) and substitute $K=\dfrac{\sin \left( \alpha \right)\sin \left( \beta \right)}{\cos \left( \alpha \right)\cos \left( \beta \right)}$ in it will result into,
$\begin{align}
  & \dfrac{1+K}{1-K}=\dfrac{1+\dfrac{\sin \left( \alpha \right)\sin \left( \beta \right)}{\cos \left( \alpha \right)\cos \left( \beta \right)}}{1-\dfrac{\sin \left( \alpha \right)\sin \left( \beta \right)}{\cos \left( \alpha \right)\cos \left( \beta \right)}} \\
 & \Rightarrow \dfrac{1+K}{1-K}=\dfrac{\dfrac{\cos \left( \alpha \right)\cos \left( \beta \right)+\sin \left( \alpha \right)\sin \left( \beta \right)}{\cos \left( \alpha \right)\cos \left( \beta \right)}}{\dfrac{\cos \left( \alpha \right)\cos \left( \beta \right)-\sin \left( \alpha \right)\sin \left( \beta \right)}{\cos \left( \alpha \right)\cos \left( \beta \right)}} \\
 & \Rightarrow \dfrac{1+K}{1-K}=\dfrac{\cos \left( \alpha \right)\cos \left( \beta \right)+\sin \left( \alpha \right)\sin \left( \beta \right)}{\cos \left( \alpha \right)\cos \left( \beta \right)-\sin \left( \alpha \right)\sin \left( \beta \right)} \\
\end{align}$
By the formulas $\cos \left( \alpha -\beta \right)=\cos \left( \alpha \right)\cos \left( \beta \right)+\sin \left( \alpha \right)\sin \left( \beta \right)$ and $\cos \left( \alpha +\beta \right)=\cos \left( \alpha \right)\cos \left( \beta \right)-\sin \left( \alpha \right)\sin \left( \beta \right)$ we will get $\dfrac{\cos \left( \alpha -\beta \right)}{\cos \left( \alpha +\beta \right)}=\dfrac{1+K}{1-K}$ which is the required answer here. We can also divide the numerator and denominator of the equation $\dfrac{\cos \left( \alpha -\beta \right)}{\cos \left( \alpha +\beta \right)}=\dfrac{\cos \left( \alpha \right)\cos \left( \beta \right)+\sin \left( \alpha \right)\sin \left( \beta \right)}{\cos \left( \alpha \right)\cos \left( \beta \right)-\sin \left( \alpha \right)\sin \left( \beta \right)}$ by $\sin \left( \alpha \right)\sin \left( \beta \right)$ also and we will get to the same result.