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If x=n. y, n R^+ the maximum value of ^2(x-y) is equal to-

Question

If $tan x = n . tan y, n ϵ R^{+}$
the maximum value of $sec^{2} (x - y)$ is equal to-

\frac{( n + 1 ) ^{2}}{2 n}

\frac{( n + 1 ) ^{2}}{n}

\frac{( n + 1 ) ^{2}}{2}

\frac{( n + 1 ) ^{2}}{4 n}

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James Discussed

tan x = n . tan y, n ϵ R^{+}

the maximum value of

sec^{2} (x - y)

is equal to-

11 mins ago

Discuss this question LIVE

11 mins ago

Text solutionVerified

Let,

z = sec^{2} (x - y) = 1 + tan^{2} (x - y)

\Rightarrow z = 1 + (\frac{tan x - tan y}{1 + tan x tan y})^{2} = 1 + (\frac{n tan y - tan y}{1 + n tan ^{2} y})^{2}

\Rightarrow z = 1 + (n - 1)^{2} (\frac{tan y}{1 + n tan ^{2} y})^{2} = 1 + (n - 1)^{2} (\frac{t}{1 + n t ^{2}})^{2}

by putting

t = tan y

Now for max value of

z

we have

\frac{d z}{d t} = 0

\Rightarrow 2 (n - 1)^{2} (\frac{t}{1 + n t ^{2}}) (\frac{1 + n t ^{2} - t ( 2 n t )}{( 1 + n t ^{2} ) ^{2}}) = 0

\Rightarrow 1 - n t^{2} = 0 \Rightarrow t^{2} = \frac{1}{n}

Hence

z_{ma x} = 1 + (n - 1)^{2} (\frac{1/ n}{( 1 + n ( 1/ n ) ) ^{2}}) = 1 + \frac{( n - 1 ) ^{2}}{4 n} = \frac{( n + 1 ) ^{2}}{4 n}

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Question Text	If $tan x = n . tan y, n ϵ R^{+}$ the maximum value of $sec^{2} (x - y)$ is equal to-
Answer Type	Text solution:1
Upvotes	150