Question

If $\tan \left(\alpha \right)=\left(k\right)cot\left(\beta \right),$ then $\frac{\cos \left(\alpha -\beta \right)}{\cos \left(\alpha +\beta \right)}$ is equal to

• A. $\frac{1+k}{1-k}$
• B. $\frac{1-k}{1+k}$
• C. $\frac{k+1}{k-1}$
• D. $\frac{k-1}{k+1}$

Answer 1

The correct option is A

$\frac{1+k}{1-k}$

Explanation for the correct option:

Step 1: Given that,

$\begin{array}{rcl}\tan \left(\alpha \right)& =& \left(k\right)cot\left(\beta \right)\\ \frac{\tan \left(\alpha \right)}{cot\left(\beta \right)}& =& k\\ \frac{{\displaystyle \frac{\sin \left(\alpha \right)}{\cos \left(\alpha \right)}}}{{\displaystyle \frac{\sin \left(\beta \right)}{\cos \left(\beta \right)}}}& =& k\\ k& =& \frac{\sin \left(\alpha \right)\sin \left(\beta \right)}{\cos \left(\alpha \right)\cos \left(\beta \right)}...\left(1\right)\end{array}$

Step 2: Apply the formula

$\begin{array}{rcl}& =& \frac{\cos \left(\alpha -\beta \right)}{\cos \left(\alpha +\beta \right)}\left[\because \cos \left(a+b\right)=\cos \left(a\right)\cos \left(b\right)-\sin \left(a\right)\sin \left(b\right)\right]\mathrm{and}\left[\because \cos \left(\mathrm{a}-\mathrm{b}\right)=\cos \left(\mathrm{a}\right)\cos \left(\mathrm{b}\right)+\sin \left(\mathrm{a}\right)\sin \left(\mathrm{b}\right)\right]\end{array}$

$\begin{array}{rcl}& =& \frac{\cos \left(\begin{array}{r}\alpha \end{array}\right)\cos \left(\beta \right)+\sin \left(\alpha \right)\sin \left(\beta \right)}{\cos \left(\begin{array}{l}\alpha \end{array}\right)\cos \left(\beta \right)-\sin \left(\alpha \right)\sin \left(\beta \right)}\end{array}$

Step 3: On dividing numerator & denominator by $\cos \left(\alpha \right)\cos \left(\beta \right)$

$\begin{array}{rcl}\therefore \frac{\cos \left(\alpha -\beta \right)}{\cos \left(\alpha +\beta \right)}& =& \frac{{\displaystyle \frac{\cos \left(\alpha \right)\cos \left(\beta \right)+\sin \left(\alpha \right)\sin \left(\beta \right)}{\cos \left(\alpha \right)\cos \left(\beta \right)}}}{{\displaystyle \frac{\cos \left(\alpha \right)\cos \left(\beta \right)-\sin \left(\alpha \right)\sin \left(\beta \right)}{\cos \left(\alpha \right)\cos \left(\beta \right)}}}\\ & =& \frac{1+{\displaystyle \frac{\sin \left(\alpha \right)\sin \left(\beta \right)}{\cos \left(\alpha \right)\cos \left(\beta \right)}}}{1-{\displaystyle \frac{\sin \left(\alpha \right)\sin \left(\beta \right)}{\cos \left(\alpha \right)\cos \left(\beta \right)}}}\end{array}$

Step 4: From the equation $\left(1\right)$

$\therefore \begin{array}{rcl}& & \frac{\cos \left(\alpha -\beta \right)}{\cos \left(\alpha +\beta \right)}\end{array}\begin{array}{rcl}& =& \end{array}\begin{array}{rcl}& & \frac{1+k}{1-k}\end{array}$

Hence, the correct option is (A).