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Question: If \(\dfrac{\tan \left( \alpha -\beta \right)}{\tan \alpha }+\dfrac{{{\sin }^{2}}\gamma }{{{\sin }^{...

If tan⁡(α−β)tan⁡α+sin⁡2γsin⁡2α=1,\dfrac{\tan \left( \alpha -\beta \right)}{\tan \alpha }+\dfrac{{{\sin }^{2}}\gamma }{{{\sin }^{2}}\alpha }=1, then prove that tan⁡γ\tan \gamma is geometric mean of tan⁡α\tan \alpha and tan⁡β\tan \beta , i.e., tan⁡αtan⁡β=tan⁡2γ\tan \alpha \tan \beta ={{\tan }^{2}}\gamma .

Explanation

Solution

Hint: Use the identities such astan⁡A=sin⁡Acos⁡A,sin(A−B)=sin Acos⁡Bcos⁡Asin⁡B,cos(A+B)=cos⁡Acos⁡B−sin⁡Asin⁡B\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⁡θ=1tan⁡θ\cot \theta =\dfrac{1}{\tan \theta } to get the desired result.

“Complete step-by-step answer:”

We are given the following equation,
tan⁡(α−β)tan⁡α+sin⁡2γsin⁡2α=1\dfrac{\tan \left( \alpha -\beta \right)}{\tan \alpha }+\dfrac{{{\sin }^{2}}\gamma }{{{\sin }^{2}}\alpha }=1………….(i)
Now by moving tan⁡(α−β)tan⁡α\dfrac{\tan \left( \alpha -\beta \right)}{\tan \alpha } from left hand side to right hand side of the equation (i) we get,
sin⁡2γsin⁡2α=1−tan⁡(α−β)tan⁡α\dfrac{{{\sin }^{2}}\gamma }{{{\sin }^{2}}\alpha }=1-\dfrac{\tan \left( \alpha -\beta \right)}{\tan \alpha }………….(ii)
Now we are using the formula,
tan⁡α=sin⁡αcos⁡α\tan \alpha =\dfrac{\sin \alpha }{\cos \alpha } and tan⁡(α−β)=sin⁡(α−β)cos⁡(α−β)\tan \left( \alpha -\beta \right)=\dfrac{\sin \left( \alpha -\beta \right)}{\cos \left( \alpha -\beta \right)}
And substituting in equation (ii) we get,
sin⁡2γsin⁡2α=1−cos⁡αsin⁡(α−β)sin⁡αcos⁡(α−β)\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,
sin⁡2γsin⁡2α=sin⁡αcos⁡(α−β)−cos⁡α(α−β)sin⁡αcos⁡(α−β)\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(A−B)=sin Acos⁡Bcos⁡Asin⁡Bsin\left( A-B \right)\text{=sin }A\cos B\cos A\sin B and we will apply by replacing A by α\alpha and B by (α−β)\left( \alpha -\beta \right). We get,
sin⁡2γsin⁡2α=sin⁡βsin⁡αcos⁡(α−β)\dfrac{{{\sin }^{2}}\gamma }{{{\sin }^{2}}\alpha }=\dfrac{\sin \beta }{\sin \alpha \cos \left( \alpha -\beta \right)}…………..(v)
On further simplification by multiplying sin⁡α\sin \alpha to both the sides of equation (v) we get,
sin⁡2γsin⁡2α=sin⁡βcos⁡(α−β)\dfrac{{{\sin }^{2}}\gamma }{{{\sin }^{2}}\alpha }=\dfrac{\sin \beta }{\cos \left( \alpha -\beta \right)}
Now by using cross multiplication we get,
sin⁡2γcos⁡(α−β)=sin⁡αsin⁡β{{\sin }^{2}}\gamma \cos \left( \alpha -\beta \right)=\sin \alpha \sin \beta ……………(vi)
Now we will use the identity cos⁡(α−β)=cos⁡αcos⁡β+sin⁡αsin⁡β\cos \left( \alpha -\beta \right)=\cos \alpha \cos \beta +\sin \alpha \sin \beta in equation (vi), we get,
sin2γ(cos⁡αcos⁡β+sin⁡αsin⁡β)=sin⁡αsin⁡βsi{{n}^{2}}\gamma \left( \cos \alpha \cos \beta +\sin \alpha \sin \beta \right)=\sin \alpha \sin \beta …………(vii)
Now dividing (sin⁡αsin⁡βsin⁡2γ)\left( \sin \alpha \sin \beta {{\sin }^{2}}\gamma \right) throughout whole equation (vii) we get,
sin⁡2γ(cos⁡αcos⁡β+sin⁡αsin⁡β)sin⁡2γsin⁡αsin⁡β=sin⁡αsin⁡βsin⁡2γsin⁡αsin⁡β\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 }
⇒(cos⁡αcos⁡βsin⁡αsin⁡β+1)=(1sin⁡2γ)\Rightarrow \left( \dfrac{\cos \alpha \cos \beta }{\sin \alpha \sin \beta }+1 \right)=\left( \dfrac{1}{{{\sin }^{2}}\gamma } \right)
⇒\Rightarrow cot⁡αcotβ+1=cosec2γ\cot \alpha cot\beta +1=cose{{c}^{2}}\gamma ……………. (viii)
Now let’s use the identity cos⁡ec2γ=1+cot⁡2γ\cos e{{c}^{2}}\gamma =1+{{\cot }^{2}}\gamma in equation (viii), we get
1+cot⁡αcot⁡β+1=1+cot⁡2γ1+\cot \alpha \cot \beta +1=1+{{\cot }^{2}}\gamma
⇒cot⁡αcot⁡β=cot⁡2γ\Rightarrow \cot \alpha \cot \beta ={{\cot }^{2}}\gamma ……………… (ix)
Now in equation (ix) we will use the identity
cot⁡θ=1tan⁡θ\cot \theta =\dfrac{1}{\tan \theta } whereθ\theta can be replace by α,β,γ\alpha ,\beta ,\gamma we get,
1tan⁡α.1tan⁡β=1tan⁡2γ ∴tan⁡αtan⁡β=tan⁡2γ \begin{aligned} & \dfrac{1}{\tan \alpha }.\dfrac{1}{\tan \beta }=\dfrac{1}{{{\tan }^{2}}\gamma } \\\ & \therefore \tan \alpha \tan \beta ={{\tan }^{2}}\gamma \\\ \end{aligned}
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⁡(α−β)\tan \left( \alpha -\beta \right). This way will be lengthy and tedious.