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Grade 11Trigonometry

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

Shristy Kashyap

9 Years agoGrade 11

1 Answer

Deep Chand

8 Years ago

We know that

$tan(x-y)=\frac{tanx-tany}{1+tanxtany}...........(1)$

We have given that

$tanx=ntany...........(2)$

Divide by $tany$ N^rand D^r 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

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