If f(x) = cos^2 x + sec^2 x,then its value is | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(D) Given the function $f(x) = \cos^2 x + \sec^2 x$.We can rewrite the expression by completing the square:$f(x) = (\cos x - \sec x)^2 + 2 \cos x \sec

Vedclass · Accepted Answer

$f(x) \geq 2$

Vedclass · Answer

(D) Given the function $f(x) = \cos^2 x + \sec^2 x$.We can rewrite the expression by completing the square:$f(x) = (\cos x - \sec x)^2 + 2 \cos x \sec x$Since $\cos x \cdot \sec x = 1$,the expression becomes:$f(x) = (\cos x - \sec x)^2 + 2(1)$$f(x) = (\cos x - \sec x)^2 + 2$Since the square of any real number is always non-negative,i.e.,$(\cos x - \sec x)^2 \geq 0$,it follows that:$f(x) \geq 0 + 2$$f(x) \geq 2$Therefore,the value of the function is always greater than or equal to $2$.

If $f(x) = \cos^2 x + \sec^2 x$,then its value is always:

Similar Questions

Let $A = \{ \theta : 2\cos^2 \theta + \sin \theta \le 2 \}$ and $B = \{ \theta : \frac{\pi}{2} \le \theta \le \frac{3\pi}{2} \}$,then $A \cap B$ is

$\cos \alpha \sin (\beta - \gamma ) + \cos \beta \sin (\gamma - \alpha ) + \cos \gamma \sin (\alpha - \beta ) = $

If $\tan \theta \cdot \cos 60^{\circ} = \frac{\sqrt{3}}{2}$,then the value of $\sin (\theta - 15^{\circ})$ is

If $A > 0, B > 0$ and $A + B = \frac{\pi}{6}$,then the minimum value of $\tan A + \tan B$ is

If $\tan A = \frac{1 - \cos B}{\sin B}$,find $\tan 2A$ in terms of $\tan B$ and show that:

Vedclass Test Series

Exam Paper Generator

Online Exam Module