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 $ .
Answer
380.4k+ views
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.
Recently Updated Pages
Basicity of sulphurous acid and sulphuric acid are

Why should electric field lines never cross each other class 12 physics CBSE

An electrostatic field line is a continuous curve That class 12 physics CBSE

What are the measures one has to take to prevent contracting class 12 biology CBSE

Suggest some methods to assist infertile couples to class 12 biology CBSE

Amniocentesis for sex determination is banned in our class 12 biology CBSE

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

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

What is pollution? How many types of pollution? Define it

Scroll valve is present in a Respiratory system of class 11 biology CBSE

What is BLO What is the full form of BLO class 8 social science CBSE

is known as the Land of the Rising Sun A Japan B Norway class 8 social science CBSE

Give 10 examples for herbs , shrubs , climbers , creepers

Write a letter to the principal requesting him to grant class 10 english CBSE

Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE
