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If tanx=ntany then maximum value of tan^2(x-y) is equal to what

Shristy Kashyap , 8 Years ago
Grade 11
anser 1 Answers
Deep Chand

Last Activity: 7 Years ago

We know that 
tan(x-y)=\frac{tanx-tany}{1+tanxtany}...........(1)
We have given that 
tanx=ntany...........(2)
Divide by  tany  Nr and Dr on RHS  in equation (1)
We get 
tan(x-y)=\frac{\frac{tanx}{tany}-1}{\frac{1}{tany}+tanx}
From equation (2) we get
tan(x-y)=\frac{n-1}{\frac{1}{tany}+ntany}
For tan(x-y) will maximum if \frac{1}{tany}+ntany  should be minimum, then
We know that 
A.M. \geq G.M.
\frac{\frac{1}{tany}+ntany}{2}\geq \sqrt{\frac{1}{tany}*ntany}
\frac{\frac{1}{tany}+ntany}{2}\geq \sqrt{n}
{\frac{1}{tany}+ntany}\geq 2\sqrt{n}
Hence 
tan(x-y)=\frac{n-1}{2\sqrt{n}}
Squaring both side we get
tan^2(x-y)=\frac{(n-1)^2}{4n}
Proved
DEEP CHAND SAROJ
B.Tech ( IIT-DELHI)
         
 

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