Equation {sec ^2}θ = frac{{4xy}}{{{{(x + y)}^2}}} is only possible when

English

Gujarati

Hindi

3.Trigonometrical Ratios, Functions and Identities

medium

The equation ${\sec ^2}\theta = \frac{{4xy}}{{{{(x + y)}^2}}}$ is only possible when

A

$x = y$

B

$x < y$

C

$x > y$

D

None of these

(IIT-1966)

Solution

(a) Since ${\cos ^2}\theta \le 1$

${\sec ^2}\theta = \frac{{4xy}}{{{{(x + y)}^2}}} \ge 1 $

$\Rightarrow 4xy \ge {(x + y)^2} $

$\Rightarrow {(x – y)^2} \le 0$

It is possible only when $x = y$, .$(x,\,y \in R)$

Standard 11

Mathematics

Share

Similar Questions

The value of $\frac{{\cot 54^\circ }}{{\tan 36^\circ }} + \frac{{\tan 20^\circ }}{{\cot 70^\circ }}$ is

easy

Prove that: $\sin 3 x+\sin 2 x-\sin x=4 \sin x \cos \frac{x}{2} \cos \frac{3 x}{2}$

hard

If $\sin A,\cos A$ and $\tan A$ are in $G.P.$, then ${\cos ^3}A + {\cos ^2}A$ is equal to

medium

Find $\sin \frac{x}{2}, \cos \frac{x}{2}$ and $\tan \frac{x}{2}$ for $\sin x=\frac{1}{4}, x$ in quadrant $II$

hard

If $\cos \theta = \frac{1}{2}\left( {x + \frac{1}{x}} \right)$, then $\frac{1}{2}\left( {{x^2} + \frac{1}{{{x^2}}}} \right) = $

easy