First slide

Compound angles

Question

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

Moderate

A

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

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

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

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

Solution

$\begin{array}{l} \tan x = ntan y, \cos (x - y) = \cos xcos y + \sin xsin y \\ \Rightarrow \cos (x - y) = \cos xcos y (1 + \tan xtan y) \\ = \cos xcos y (1 + {ntan}^{2} y) \\ \Rightarrow \sec^{2} (x - y) = \frac{\sec^{2} {xsec}^{2} y}{{(1 + {ntan}^{2} y)}^{2}} \end{array}$
$\begin{array}{l} = \frac{(1 + \tan^{2} x) (1 + \tan^{2} y)}{{(1 + {ntan}^{2} y)}^{2}} \\ = \frac{(1 + n^{2} \tan^{2} y) (1 + \tan^{2} y)}{{(1 + {ntan}^{2} y)}^{2}} \\ = 1 + \frac{(n - 1)^{2} \tan^{2} y}{{(1 + {ntan}^{2} y)}^{2}} \end{array}$
$Now, {(\frac{1 + {ntan}^{2} y}{2})}^{2} \geq {ntan}^{2} y (∵ A.M. \geq G.M.)$
$\begin{array}{l} or \frac{\tan^{2} y}{{(1 + {ntan}^{2} y)}^{2}} \leq \frac{1}{4 n} \\ \Rightarrow \sec^{2} (x - y) \leq 1 + \frac{(n - 1)^{2}}{4 n} = \frac{(n + 1)^{2}}{4 n} \end{array}$

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