Question

If $\frac{\cos x}{\cos (x-2y)}=\lambda$, then $\tan (x-y)\tan y$ is equal to

• A. $\frac{(1+\lambda )}{(1–\lambda )}$
• B. $\frac{(1-\lambda )}{(1+\lambda )}$
• C. $\frac{\lambda }{(1+\lambda )}$
• D. $\frac{\lambda }{(1–\lambda )}$

Answer 1

The correct option is B

$\frac{(1-\lambda )}{(1+\lambda )}$

Explanation for the correct option:

Step 1. Find the value of $\tan (x-y)\tan y$:

Given, $\frac{\cos x}{\cos (x-2y)}=\lambda$

Now,

$\begin{array}{rcl}\tan (x–y)\tan y& =& \left[\frac{\sin (x–y)}{\cos (x–y)}\right]\left(\frac{\sin y}{\cos y}\right)\\ & =& \frac{\left[2\sin (x–y)\sin y\right]}{\left[2\cos (x–y)\cos y\right]}\end{array}$

Step 2. Use the formulas $\mathbf{2}\mathbf{}\mathit{s}\mathit{i}\mathit{n}\mathbf{}\mathit{C}\mathbf{}\mathit{s}\mathit{i}\mathit{n}\mathbf{}\mathit{D}\mathbf{}\mathbf{=}\mathbf{}\mathit{c}\mathit{o}\mathit{s}\mathbf{(}\mathit{C}\mathbf{}\mathbf{–}\mathbf{}\mathit{D}\mathbf{)}\mathbf{}\mathbf{–}\mathbf{}\mathit{c}\mathit{o}\mathit{s}\mathbf{(}\mathit{C}\mathbf{}\mathbf{+}\mathbf{}\mathit{D}\mathbf{)}\mathbf{}$and $\mathbf{2}\mathbf{}\mathit{c}\mathit{o}\mathit{s}\mathbf{}\mathit{C}\mathbf{}\mathit{c}\mathit{o}\mathit{s}\mathbf{}\mathit{D}\mathbf{}\mathbf{=}\mathbf{}\mathit{c}\mathit{o}\mathit{s}\mathbf{(}\mathit{C}\mathbf{}\mathbf{+}\mathbf{}\mathit{D}\mathbf{)}\mathbf{}\mathbf{+}\mathbf{}\mathit{c}\mathit{o}\mathit{s}\mathbf{(}\mathit{C}\mathbf{}\mathbf{–}\mathbf{}\mathit{D}\mathbf{)}$

$$\begin{gathered} =\frac{\cos (x–y–y)–\cos (x–y+y)}{\cos (x–y+y)+\cos (x–y–y)} \\ =\frac{\cos (x–2y)–\cos x}{\cos x+\cos (x–2y)} \end{gathered}$$

Step 3. Take and cancel $\cos (x–2y)$ from numerator and denominator, we get

$$\begin{gathered} =\frac{1–\left[{\displaystyle \frac{\cos x}{\cos (x–2y)}}\right]}{\left[{\displaystyle \frac{\cos x}{\cos (x–2y)}}\right]+1} \\ =\frac{(1–\lambda )}{(\lambda +1)} \\ =\frac{\mathbf{(}\mathbf{1}\mathbf{}\mathbf{–}\mathbf{}\mathbf{\lambda }\mathbf{)}}{\mathbf{(}\mathbf{1}\mathbf{}\mathbf{+}\mathbf{}\mathbf{\lambda }\mathbf{)}} \end{gathered}$$

Hence, Option ‘B’ is Correct.