Question

if tan?=ntan? ,then the maximum value of tan^2(?-?) is equal to ??
Answer:- [(n-1)^2]/4n
My actual question :- how to obtain this answer ??

Answer 1

Dear student
$$\begin{gathered} Given:\tan \theta =n\tan \varphi \\ Now, \\ \cos \left(\theta -\varphi \right)=\cos \theta \cos \varphi -\sin \theta \sin \varphi \\ cos\left(\theta -\varphi \right)=\cos \theta \cos \varphi (1+\tan \theta \tan \varphi ) \\ cos\left(\theta -\varphi \right)=\cos \theta \cos \varphi (1+n{\tan }^2\varphi ) \\ Now, \\ se{c}^2\left(\theta -\varphi \right)=\frac{se{c}^2\theta se{c}^2\varphi }{{\left(1+n{\tan }^2\varphi \right)}^2} \\ =\frac{\left(1+ta{n}^2\theta \right)\left(1+ta{n}^2\varphi \right)}{{\left(1+nta{n}^2\varphi \right)}^2} \\ =1+\frac{{\left(n-1\right)}^2{\tan }^2\varphi }{{\left(1+nta{n}^2\varphi \right)}^2} \\ Now, \\ {\left(\frac{1+nta{n}^2\varphi }{2}\right)}^2\ge n{\tan }^2\varphi \\ \Rightarrow \frac{ta{n}^2\varphi }{{\left(1+nta{n}^2\varphi \right)}^2}\le \frac{1}{4n} \\ \Rightarrow se{c}^2\left(\theta -\varphi \right)\le 1+\frac{{\left(n-1\right)}^2}{4n}=\frac{{\left(n+1\right)}^2}{4n} \\ or \\ {\tan }^2\left(\theta -\varphi \right)-1\le 1+\frac{{\left(n-1\right)}^2}{4n}-1=\frac{{\left(n-1\right)}^2}{4n} \end{gathered}$$
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