Question

If $\tan \beta = \dfrac{{\tan \alpha + \tan \gamma }}{{1 + \tan \alpha \tan \gamma }}$prove that \sin 2\beta $$$$ = \dfrac{{\sin 2\alpha + \sin 2\gamma }}{{1 + \sin 2\alpha \sin 2\gamma }}

Answer 1

Hint : So, here we need to first solve for $\tan \beta$ and then we have to prove for $\sin 2\beta$. In order to prove, we need to know some trigonometric formulas are $\tan \theta = \dfrac{{\sin \theta }}{{\cos \theta }}$,
$\sin (A + B) = \sin A\operatorname{Cos} B + \operatorname{Cos} A\operatorname{Sin} B$,
$\cos (A - B) = \operatorname{Cos} A\operatorname{Cos} B - \operatorname{Sin} A\operatorname{Sin} B$,
By using all trigonometric identities to get the required solution.

Complete step-by-step answer :
We are given, $\tan \beta = \dfrac{{\tan \alpha + \tan \gamma }}{{1 + \tan \alpha \tan \gamma }}$
First, comparing the trigonometric identity with the above function, then
Now, we know the formula, $\tan \theta = \dfrac{{\sin \theta }}{{\cos \theta }}$.
$= \dfrac{{\dfrac{{\sin \alpha }}{{\cos \alpha }} + \dfrac{{\sin \gamma }}{{\cos \gamma }}}}{{1 + \left( {\dfrac{{\sin \alpha }}{{\cos \alpha }}} \right)\left( {\dfrac{{\sin \gamma }}{{\cos \gamma }}} \right)}}$
Now, we need to cross multiply on both numerator and denominator, we will get;
$= \dfrac{{\sin \alpha \cos \gamma + \cos \alpha \sin \gamma }}{{1 + \left( {\dfrac{{\sin \alpha }}{{\cos \alpha }}} \right)\left( {\dfrac{{\sin \gamma }}{{\cos \gamma }}} \right)}}$
$= \dfrac{{\sin \alpha \cos \gamma + \cos \alpha \sin \gamma }}{{1 + \left( {\dfrac{{\sin \alpha \sin \gamma }}{{\cos \alpha \cos \gamma }}} \right)}}$
$= \dfrac{{\sin \alpha \cos \gamma + \cos \alpha \sin \gamma }}{{\cos \alpha \cos \gamma + \sin \alpha \sin \gamma }}$
Now, we know the formula of trigonometric function, we have
$\sin (A + B) = \sin A\operatorname{Cos} B + \operatorname{Cos} A\operatorname{Sin} B$,
$\cos (A - B) = \operatorname{Cos} A\operatorname{Cos} B - \operatorname{Sin} A\operatorname{Sin} B$,
So let us substitute the formulas in the above equation, we get;
$= \dfrac{{\sin (\alpha + \gamma )}}{{\cos (\alpha - \gamma )}}$
Therefore, $\tan \beta = \dfrac{{\sin (\alpha + \gamma )}}{{\cos (\alpha - \gamma )}}$ ---------(1)
Now, let us take $\sin 2\beta$, which is given in the question.
$\sin 2\beta = \dfrac{{\sin 2\alpha + \sin 2\gamma }}{{1 + \sin 2\alpha \sin 2\gamma }}$
We know the formula for $\sin 2\beta$, which we will use, and it is given as:
$\Rightarrow \sin 2\beta = \dfrac{{2\tan \beta }}{{1 + {{\tan }^2}\beta }}$
As we mentioned earlier the formula for $\tan \theta = \dfrac{{\sin \theta }}{{\cos \theta }}$, so let us substitute in the above mentioned equation, we get;
$= \dfrac{{2\dfrac{{\sin (\alpha + \gamma )}}{{\cos (\alpha - \gamma )}}}}{{1 + {{\left( {\dfrac{{\sin (\alpha + \gamma )}}{{\cos (\alpha - \gamma )}}} \right)}^2}}}$
Now, we need to again cross multiply for the above equation of denominator then we get,
$= \dfrac{{2\dfrac{{\sin (\alpha + \gamma )}}{{\cos (\alpha - \gamma )}}}}{{\dfrac{{{{(\cos (\alpha - \gamma ))}^2} + {{(\sin (\alpha + \gamma ))}^2}}}{{{{(\cos (\alpha - \gamma ))}^2}}}}}$
On simplifying, we get
$= \dfrac{{2\sin (\alpha + \gamma )\cos (\alpha - \gamma )}}{{{{(\cos (\alpha - \gamma ))}^2} + {{(\sin (\alpha + \gamma ))}^2}}}$
Now, in the above equation, numerator is in the form of:
$\Rightarrow 2\sin (A)\cos (B) = \sin (A + B) + \sin (A - B)$
$= \dfrac{{\sin (\alpha + \gamma + \alpha - \gamma ) + \sin (\alpha + \gamma - \alpha + \gamma )}}{{{{\cos }^2}(\alpha - \gamma ) + {{\sin }^2}(\alpha + \gamma )}}$
$= \dfrac{{\sin (2\alpha ) + \sin (2\gamma )}}{{{{\cos }^2}(\alpha - \gamma ) + {{\sin }^2}(\alpha + \gamma )}}$
Now, let us multiply both numerator and denominator by 2, then:
$= \dfrac{{2[\sin (2\alpha ) + \sin (2\gamma )]}}{{2[{{\cos }^2}(\alpha - \gamma ) + {{\sin }^2}(\alpha + \gamma )]}}$
$= \dfrac{{2[\sin (2\alpha ) + \sin (2\gamma )]}}{{2{{\cos }^2}(\alpha - \gamma ) + 2{{\sin }^2}(\alpha + \gamma )}}$
We know that ,

$$\Rightarrow 2{\cos ^2}\theta = 1 + \cos 2\theta \\\ \Rightarrow 2{\sin ^2}\theta = 1 - \cos 2\theta \;$$

$\Rightarrow \cos 2\theta = {\cos ^2}\theta - {\sin ^2}\theta$
$\Rightarrow \cos 2\theta = 2{\cos ^2}\theta - 1$
$\Rightarrow \cos 2\theta = 1 - 2{\sin ^2}\theta$
Now, by using the above mentioned formula, we will substitute in the aforementioned equation, then we will get;
$= \dfrac{{2[\sin (2\alpha ) + \sin (2\gamma )]}}{{1 + \cos (2\alpha - 2\gamma ) + 1 - \cos (2\alpha + 2\gamma )}}$
$= \dfrac{{2[\sin (2\alpha ) + \sin (2\gamma )]}}{{2 + \cos (2\alpha - 2\gamma ) - \cos (2\alpha - 2\gamma )}}$
Now, the denominator in the above equation is in the form of:
$\Rightarrow \cos (A - B) - cos(A + B) = \cos A\cos B + \sin A\sin B - \cos A\cos B + \sin A\sin B$
$\Rightarrow \cos (A - B) - cos(A + B) = 2\sin A\sin B$
Now, by applying the above formula in the equation, we get:
$= \dfrac{{2[\sin (2\alpha ) + \sin (2\gamma )]}}{{2 + 2\sin 2\alpha \sin 2\gamma }}$
Now let us take two as common both in numerator and denominator, which gets cancelled,
$= \dfrac{{\sin 2\alpha + \sin 2\gamma }}{{1 + \sin 2\alpha \sin 2\gamma }}$
Therefore, L.H.S = R.H.S
Hence, Proved.

Note : It is must that most of the trigonometric equations always rely on formulas which are mentioned above and if we follow that then we will be able to achieve our solution. We need to prove the given problem by using trigonometric identity and substitution methods.