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
•
Total views: 304.5k
•
Views today: 2.84k
Answer
304.5k+ 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
If ab and c are unit vectors then left ab2 right+bc2+ca2 class 12 maths JEE_Main

A rod AB of length 4 units moves horizontally when class 11 maths JEE_Main

Evaluate the value of intlimits0pi cos 3xdx A 0 B 1 class 12 maths JEE_Main

Which of the following is correct 1 nleft S cup T right class 10 maths JEE_Main

What is the area of the triangle with vertices Aleft class 11 maths JEE_Main

KCN reacts readily to give a cyanide with A Ethyl alcohol class 12 chemistry JEE_Main

Trending doubts
What was the capital of Kanishka A Mathura B Purushapura class 7 social studies CBSE

Difference Between Plant Cell and Animal Cell

Write an application to the principal requesting five class 10 english CBSE

Ray optics is valid when characteristic dimensions class 12 physics CBSE

Give 10 examples for herbs , shrubs , climbers , creepers

Tropic of Cancer passes through how many states? Name them.

Write the 6 fundamental rights of India and explain in detail

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

Name the Largest and the Smallest Cell in the Human Body ?
