Question

If $\cos \theta =\cos \alpha \cos \beta$,then $\tan \frac{(\theta +\alpha )}{2}\tan \frac{(\theta -\alpha )}{2}$ is equal to

• A. ${\tan }^2\frac{\alpha }{2}$
• B. ${\tan }^2\frac{\beta }{2}$
• C. ${\tan }^2\frac{\theta }{2}$
• D. $co{t}^2\frac{\beta }{2}$

Answer 1

The correct option is B

${\tan }^2\frac{\beta }{2}$

Explanation for the correct option:

Find the value of $\mathit{t}\mathit{a}\mathit{n}\frac{\mathbf{(}\mathbf{\theta }\mathbf{}\mathbf{+}\mathbf{}\mathbf{\alpha }\mathbf{)}}{\mathbf{2}}\mathbf{}\mathit{t}\mathit{a}\mathit{n}\frac{\mathbf{(}\mathbf{\theta }\mathbf{}\mathbf{-}\mathbf{}\mathbf{\alpha }\mathbf{)}}{\mathbf{2}}\mathbf{:}$

Given, $\cos \theta =\cos \alpha \cos \beta$

$\Rightarrow$ $\frac{\cos \theta }{\cos \alpha }=\cos \beta$

Apply componendo and dividendo

$\Rightarrow$$\frac{\cos \theta +\cos \alpha }{\cos \theta -\cos \alpha }=\frac{\cos \beta +1}{\cos \beta -1}$

Using formulae,

$$\begin{gathered} \mathit{c}\mathit{o}\mathit{s}\mathbf{}\mathit{C}\mathbf{}\mathbf{+}\mathbf{}\mathit{c}\mathit{o}\mathit{s}\mathbf{}\mathit{D}\mathbf{}\mathbf{=}\mathbf{}\mathbf{2}\mathbf{}\mathit{c}\mathit{o}\mathit{s}\left(\frac{C+D}{2}\right)\mathbf{}\mathit{c}\mathit{o}\mathit{s}\left(\frac{C-D}{2}\right) \\ \mathit{c}\mathit{o}\mathit{s}\mathbf{}\mathit{C}\mathbf{}\mathbf{-}\mathbf{}\mathit{c}\mathit{o}\mathit{s}\mathbf{}\mathit{D}\mathbf{}\mathbf{=}\mathbf{}\mathbf{-}\mathbf{2}\mathbf{}\mathit{s}\mathit{i}\mathit{n}\left(\frac{C+D}{2}\right)\mathbf{}\mathit{s}\mathit{i}\mathit{n}\left(\frac{C-D}{2}\right) \\ \mathbf{1}\mathbf{}\mathbf{+}\mathbf{}\mathit{c}\mathit{o}\mathit{s}\mathbf{}\mathit{\beta }\mathbf{}\mathbf{=}\mathbf{}\mathbf{2}\mathbf{}\mathit{c}\mathit{o}{\mathit{s}}^{\mathbf{2}}\left(\frac{\beta }{2}\right) \\ \mathit{c}\mathit{o}\mathit{s}\mathbf{}\mathit{\beta }\mathbf{}\mathbf{-}\mathbf{}\mathbf{1}\mathbf{}\mathbf{=}\mathbf{}\mathbf{-}\mathbf{2}\mathbf{}\mathit{s}\mathit{i}{\mathit{n}}^{\mathbf{2}}\left(\frac{\beta }{2}\right) \end{gathered}$$

$\Rightarrow$$\frac{2\cos \left({\displaystyle \frac{\theta +\alpha }{2}}\right)\cos \left(\frac{\theta -\alpha }{2}\right)}{-2\sin \left(\frac{\theta +\alpha }{2}\right)\sin \left(\frac{\theta -\alpha }{2}\right)}=\frac{2{\cos }^2\left({\displaystyle \frac{\beta }{2}}\right)}{-2{\sin }^2\left(\frac{\beta }{2}\right)}$

Reciprocal above equation

$\Rightarrow$ $\frac{\sin \left({\displaystyle \frac{\theta +\alpha }{2}}\right)\sin \left(\frac{\theta -\alpha }{2}\right)}{\cos \left(\frac{\theta +\alpha }{2}\right)\cos \left(\frac{\theta -\alpha }{2}\right)}=\frac{{\sin }^2\left({\displaystyle \frac{\beta }{2}}\right)}{{\cos }^2\left(\frac{\beta }{2}\right)}$

$\mathbf{\therefore }\mathit{t}\mathit{a}\mathit{n}\frac{\mathbf{(}\mathbf{\theta }\mathbf{}\mathbf{+}\mathbf{}\mathbf{\alpha }\mathbf{)}}{\mathbf{2}}\mathbf{}\mathit{t}\mathit{a}\mathit{n}\frac{\mathbf{(}\mathbf{\theta }\mathbf{}\mathbf{-}\mathbf{}\mathbf{\alpha }\mathbf{)}}{\mathbf{2}}$$=\mathit{t}\mathit{a}{\mathit{n}}^{\mathbf{2}}\mathbf{\left(}\frac{\mathbf{\beta }}{\mathbf{2}}\mathbf{\right)}$

Hence, Option ‘B’ is Correct.