Answer

Verified

365.1k+ views

**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.

Recently Updated Pages

Mark and label the given geoinformation on the outline class 11 social science CBSE

When people say No pun intended what does that mea class 8 english CBSE

Name the states which share their boundary with Indias class 9 social science CBSE

Give an account of the Northern Plains of India class 9 social science CBSE

Change the following sentences into negative and interrogative class 10 english CBSE

Advantages and disadvantages of science

Trending doubts

Bimbisara was the founder of dynasty A Nanda B Haryanka class 6 social science CBSE

Which are the Top 10 Largest Countries of the World?

Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE

Differentiate between homogeneous and heterogeneous class 12 chemistry CBSE

10 examples of evaporation in daily life with explanations

Fill the blanks with the suitable prepositions 1 The class 9 english CBSE

Give 10 examples for herbs , shrubs , climbers , creepers

How do you graph the function fx 4x class 9 maths CBSE

Difference Between Plant Cell and Animal Cell