# If $\dfrac{\tan \left( \alpha -\beta \right)}{\tan \alpha }+\dfrac{{{\sin }^{2}}\gamma }{{{\sin }^{2}}\alpha }=1,$ then prove that $\tan \gamma $ is geometric mean of $\tan \alpha $ and $\tan \beta $ , i.e., $\tan \alpha \tan \beta ={{\tan }^{2}}\gamma $ .

Last updated date: 18th Mar 2023

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Answer

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Hint: Use the identities such as\[\tan A=\dfrac{\sin A}{\cos A},sin\left( A-B \right)\text{=sin }A\cos B\cos A\sin B,cos\left( A+B \right)=\cos A\cos B-\sin A\sin B\] in the question properly and wisely. Also try to simplify it whenever possible. At first try to convert all the terms in tan ratios to respective sin and cos ratios and after doing necessary calculation try to change the terms in cot ratios and finally use identity $\cot \theta =\dfrac{1}{\tan \theta }$ to get the desired result.

“Complete step-by-step answer:”

We are given the following equation,

$\dfrac{\tan \left( \alpha -\beta \right)}{\tan \alpha }+\dfrac{{{\sin }^{2}}\gamma }{{{\sin }^{2}}\alpha }=1$………….(i)

Now by moving $\dfrac{\tan \left( \alpha -\beta \right)}{\tan \alpha }$ from left hand side to right hand side of the equation (i) we get,

$\dfrac{{{\sin }^{2}}\gamma }{{{\sin }^{2}}\alpha }=1-\dfrac{\tan \left( \alpha -\beta \right)}{\tan \alpha }$………….(ii)

Now we are using the formula,

\[\tan \alpha =\dfrac{\sin \alpha }{\cos \alpha }\] and $\tan \left( \alpha -\beta \right)=\dfrac{\sin \left( \alpha -\beta \right)}{\cos \left( \alpha -\beta \right)}$

And substituting in equation (ii) we get,

$\dfrac{{{\sin }^{2}}\gamma }{{{\sin }^{2}}\alpha }=1-\dfrac{\cos \alpha \sin \left( \alpha -\beta \right)}{\sin \alpha \cos \left( \alpha -\beta \right)}$…………..(iii)

Now by taking LCM and further simplifying in right hand side of equation (iii) we get,

$\dfrac{{{\sin }^{2}}\gamma }{{{\sin }^{2}}\alpha }=\dfrac{\sin \alpha \cos \left( \alpha -\beta \right)-\cos \alpha \left( \alpha -\beta \right)}{\sin \alpha \cos \left( \alpha -\beta \right)}$…………….(iv)

Now in the equation (iv) we will use the identity \[sin\left( A-B \right)\text{=sin }A\cos B\cos A\sin B\] and we will apply by replacing A by $\alpha $ and B by $\left( \alpha -\beta \right)$. We get,

$\dfrac{{{\sin }^{2}}\gamma }{{{\sin }^{2}}\alpha }=\dfrac{\sin \beta }{\sin \alpha \cos \left( \alpha -\beta \right)}$…………..(v)

On further simplification by multiplying $\sin \alpha $ to both the sides of equation (v) we get,

$\dfrac{{{\sin }^{2}}\gamma }{{{\sin }^{2}}\alpha }=\dfrac{\sin \beta }{\cos \left( \alpha -\beta \right)}$

Now by using cross multiplication we get,

${{\sin }^{2}}\gamma \cos \left( \alpha -\beta \right)=\sin \alpha \sin \beta $……………(vi)

Now we will use the identity $\cos \left( \alpha -\beta \right)=\cos \alpha \cos \beta +\sin \alpha \sin \beta $ in equation (vi), we get,

$si{{n}^{2}}\gamma \left( \cos \alpha \cos \beta +\sin \alpha \sin \beta \right)=\sin \alpha \sin \beta $…………(vii)

Now dividing $\left( \sin \alpha \sin \beta {{\sin }^{2}}\gamma \right)$ throughout whole equation (vii) we get,

$\dfrac{{{\sin }^{2}}\gamma \left( \cos \alpha \cos \beta +\sin \alpha \sin \beta \right)}{{{\sin }^{2}}\gamma \sin \alpha \sin \beta }=\dfrac{\sin \alpha \sin \beta }{{{\sin }^{2}}\gamma \sin \alpha \sin \beta }$

$\Rightarrow \left( \dfrac{\cos \alpha \cos \beta }{\sin \alpha \sin \beta }+1 \right)=\left( \dfrac{1}{{{\sin }^{2}}\gamma } \right)$

$\Rightarrow $$\cot \alpha cot\beta +1=cose{{c}^{2}}\gamma $……………. (viii)

Now let’s use the identity $\cos e{{c}^{2}}\gamma =1+{{\cot }^{2}}\gamma $ in equation (viii), we get

$1+\cot \alpha \cot \beta +1=1+{{\cot }^{2}}\gamma $

$\Rightarrow \cot \alpha \cot \beta ={{\cot }^{2}}\gamma $ ……………… (ix)

Now in equation (ix) we will use the identity

$\cot \theta =\dfrac{1}{\tan \theta }$ where$\theta $ can be replace by $\alpha ,\beta ,\gamma $ we get,

$\begin{align}

& \dfrac{1}{\tan \alpha }.\dfrac{1}{\tan \beta }=\dfrac{1}{{{\tan }^{2}}\gamma } \\

& \therefore \tan \alpha \tan \beta ={{\tan }^{2}}\gamma \\

\end{align}$

Hence proved

Note: In these types of problems students should generally convert all the ‘tan’ ratios to ‘sin’ and ‘cos’ ones and then simplify using the identities.

Students should also learn the identities by heart and should be well versed how to and where to use them to get the desired results. They should also be careful while working every single step so as to avoid miscalculations.

Another approach is to expand $\tan \left( \alpha -\beta \right)$. This way will be lengthy and tedious.

“Complete step-by-step answer:”

We are given the following equation,

$\dfrac{\tan \left( \alpha -\beta \right)}{\tan \alpha }+\dfrac{{{\sin }^{2}}\gamma }{{{\sin }^{2}}\alpha }=1$………….(i)

Now by moving $\dfrac{\tan \left( \alpha -\beta \right)}{\tan \alpha }$ from left hand side to right hand side of the equation (i) we get,

$\dfrac{{{\sin }^{2}}\gamma }{{{\sin }^{2}}\alpha }=1-\dfrac{\tan \left( \alpha -\beta \right)}{\tan \alpha }$………….(ii)

Now we are using the formula,

\[\tan \alpha =\dfrac{\sin \alpha }{\cos \alpha }\] and $\tan \left( \alpha -\beta \right)=\dfrac{\sin \left( \alpha -\beta \right)}{\cos \left( \alpha -\beta \right)}$

And substituting in equation (ii) we get,

$\dfrac{{{\sin }^{2}}\gamma }{{{\sin }^{2}}\alpha }=1-\dfrac{\cos \alpha \sin \left( \alpha -\beta \right)}{\sin \alpha \cos \left( \alpha -\beta \right)}$…………..(iii)

Now by taking LCM and further simplifying in right hand side of equation (iii) we get,

$\dfrac{{{\sin }^{2}}\gamma }{{{\sin }^{2}}\alpha }=\dfrac{\sin \alpha \cos \left( \alpha -\beta \right)-\cos \alpha \left( \alpha -\beta \right)}{\sin \alpha \cos \left( \alpha -\beta \right)}$…………….(iv)

Now in the equation (iv) we will use the identity \[sin\left( A-B \right)\text{=sin }A\cos B\cos A\sin B\] and we will apply by replacing A by $\alpha $ and B by $\left( \alpha -\beta \right)$. We get,

$\dfrac{{{\sin }^{2}}\gamma }{{{\sin }^{2}}\alpha }=\dfrac{\sin \beta }{\sin \alpha \cos \left( \alpha -\beta \right)}$…………..(v)

On further simplification by multiplying $\sin \alpha $ to both the sides of equation (v) we get,

$\dfrac{{{\sin }^{2}}\gamma }{{{\sin }^{2}}\alpha }=\dfrac{\sin \beta }{\cos \left( \alpha -\beta \right)}$

Now by using cross multiplication we get,

${{\sin }^{2}}\gamma \cos \left( \alpha -\beta \right)=\sin \alpha \sin \beta $……………(vi)

Now we will use the identity $\cos \left( \alpha -\beta \right)=\cos \alpha \cos \beta +\sin \alpha \sin \beta $ in equation (vi), we get,

$si{{n}^{2}}\gamma \left( \cos \alpha \cos \beta +\sin \alpha \sin \beta \right)=\sin \alpha \sin \beta $…………(vii)

Now dividing $\left( \sin \alpha \sin \beta {{\sin }^{2}}\gamma \right)$ throughout whole equation (vii) we get,

$\dfrac{{{\sin }^{2}}\gamma \left( \cos \alpha \cos \beta +\sin \alpha \sin \beta \right)}{{{\sin }^{2}}\gamma \sin \alpha \sin \beta }=\dfrac{\sin \alpha \sin \beta }{{{\sin }^{2}}\gamma \sin \alpha \sin \beta }$

$\Rightarrow \left( \dfrac{\cos \alpha \cos \beta }{\sin \alpha \sin \beta }+1 \right)=\left( \dfrac{1}{{{\sin }^{2}}\gamma } \right)$

$\Rightarrow $$\cot \alpha cot\beta +1=cose{{c}^{2}}\gamma $……………. (viii)

Now let’s use the identity $\cos e{{c}^{2}}\gamma =1+{{\cot }^{2}}\gamma $ in equation (viii), we get

$1+\cot \alpha \cot \beta +1=1+{{\cot }^{2}}\gamma $

$\Rightarrow \cot \alpha \cot \beta ={{\cot }^{2}}\gamma $ ……………… (ix)

Now in equation (ix) we will use the identity

$\cot \theta =\dfrac{1}{\tan \theta }$ where$\theta $ can be replace by $\alpha ,\beta ,\gamma $ we get,

$\begin{align}

& \dfrac{1}{\tan \alpha }.\dfrac{1}{\tan \beta }=\dfrac{1}{{{\tan }^{2}}\gamma } \\

& \therefore \tan \alpha \tan \beta ={{\tan }^{2}}\gamma \\

\end{align}$

Hence proved

Note: In these types of problems students should generally convert all the ‘tan’ ratios to ‘sin’ and ‘cos’ ones and then simplify using the identities.

Students should also learn the identities by heart and should be well versed how to and where to use them to get the desired results. They should also be careful while working every single step so as to avoid miscalculations.

Another approach is to expand $\tan \left( \alpha -\beta \right)$. This way will be lengthy and tedious.

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