Question

If \[\cos \theta =\cos \alpha .\cos \beta \], then \[\tan \left( \dfrac{\theta +\alpha }{2} \right)\tan \left( \dfrac{\theta -\alpha }{2} \right)\] is equal to
a) \[{{\tan }^{2}}\left( \dfrac{\alpha }{2} \right)\]
b) \[{{\tan }^{2}}\left( \dfrac{\beta }{2} \right)\]
c) \[{{\tan }^{2}}\left( \dfrac{\theta }{2} \right)\]
d) \[{{\cot }^{2}}\left( \dfrac{\beta }{2} \right)\]

Answer 1

Hint: By using the \[\cos \theta =\cos \alpha .\cos \beta \], we need to find the value of \[\tan \left( \dfrac{\theta +\alpha }{2} \right)\tan \left( \dfrac{\theta -\alpha }{2} \right)\]. We know that tangent is the ratio of sine to cosine function by applying this and using some trigonometric formula we can find the value.

Formula used:
\[\cos A-\cos B=2\sin \left( \dfrac{A+B}{2} \right)\sin \left( \dfrac{B-A}{2} \right)\]
\[\cos A+\cos B=2\cos \left( \dfrac{A+B}{2} \right)\cos \left( \dfrac{A-B}{2} \right)\]
\[1-\cos A=2{{\sin }^{2}}\left( \dfrac{A}{2} \right)\]
\[1+\cos A=2{{\cos }^{2}}\left( \dfrac{A}{2} \right)\]

Step by step solution:
Given, \[\cos \theta =\cos \alpha .\cos \beta \,\,\,\,--(1)\]
Now, we have \[\tan \left( \dfrac{\theta +\alpha }{2} \right)\tan \left( \dfrac{\theta -\alpha }{2} \right)\], by definition we know that tangent is the ratio of sine to cosine.
\[\Rightarrow \tan \left( \dfrac{\theta +\alpha }{2} \right)\tan \left( \dfrac{\theta -\alpha }{2} \right)=\dfrac{\sin \left( \dfrac{\theta +\alpha }{2} \right)}{\cos \left( \dfrac{\theta +\alpha }{2} \right)}\dfrac{\sin \left( \dfrac{\theta -\alpha }{2} \right)}{\cos \left( \dfrac{\theta -\alpha }{2} \right)}\]
Multiplying 2 and dividing 2 on RHS,
\[=\dfrac{2\sin \left( \dfrac{\theta +\alpha }{2} \right)}{2\cos \left( \dfrac{\theta +\alpha }{2} \right)}\dfrac{\sin \left( \dfrac{\theta -\alpha }{2} \right)}{\cos \left( \dfrac{\theta -\alpha }{2} \right)}\]
We know that \[\cos \alpha -\cos \theta =2\sin \left( \dfrac{\theta +\alpha }{2} \right)\sin \left( \dfrac{\theta -\alpha }{2} \right)\] and \[\cos \alpha +\cos \theta =2\cos \left( \dfrac{\theta +\alpha }{2} \right)\cos \left( \dfrac{\theta -\alpha }{2} \right)\], using this we have,
\[=\dfrac{\cos \alpha -\cos \theta }{\cos \alpha +\cos \theta }\]
But from equation 1 we have \[\cos \theta =\cos \alpha .\cos \beta \],
\[=\dfrac{\cos \alpha -\cos \alpha \cos \beta }{\cos \alpha +\cos \alpha \cos \beta }\]
Taking \[\cos \alpha \] common we have
\[=\dfrac{\cos \alpha \left( 1-\cos \beta \right)}{\cos \alpha \left( 1+\cos \beta \right)}\]
\[=\dfrac{1-\cos \beta }{1+\cos \beta }\]
\[=\dfrac{2{{\sin }^{2}}\left( \dfrac{\beta }{2} \right)}{2{{\cos }^{2}}\left( \dfrac{\beta }{2} \right)}\]
 \[={{\tan }^{2}}\left( \dfrac{\beta }{2} \right)\]
Hence, \[\tan \left( \dfrac{\theta +\alpha }{2} \right)\tan \left( \dfrac{\theta -\alpha }{2} \right)={{\tan }^{2}}\left( \dfrac{\beta }{2} \right)\].

Hence, option (b) is correct.

Note: Trigonometric identities and ratios are the concept that is used to illustrate the given problem. Trigonometric functions, including variables and constants, are all those trigonometric identities. The replacement method using trigonometric functions is a typical approach used to solve this issue.