If sin x + cos x = frac{1}{5}, then tan 2x is | vedclass

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

Question

(D) Given: $\sin x + \cos x = \frac{1}{5}$Squaring both sides:$(\sin x + \cos x)^2 = (\frac{1}{5})^2$$\sin^2 x + \cos^2 x + 2 \sin x \cos x = \frac{1}

Vedclass · Accepted Answer

$\frac{24}{7}$

Vedclass · Answer

(D) Given: $\sin x + \cos x = \frac{1}{5}$Squaring both sides:$(\sin x + \cos x)^2 = (\frac{1}{5})^2$$\sin^2 x + \cos^2 x + 2 \sin x \cos x = \frac{1}{25}$Since $\sin^2 x + \cos^2 x = 1$ and $2 \sin x \cos x = \sin 2x$:$1 + \sin 2x = \frac{1}{25}$$\sin 2x = \frac{1}{25} - 1 = -\frac{24}{25}$Now,$\cos^2 2x = 1 - \sin^2 2x = 1 - (-\frac{24}{25})^2 = 1 - \frac{576}{625} = \frac{49}{625}$$\cos 2x = \pm \frac{7}{25}$Since $\sin 2x$ is negative,$2x$ lies in the $III$ or $IV$ quadrant. If we assume the standard identity approach:$	an 2x = \frac{\sin 2x}{\cos 2x} = \frac{-24/25}{\pm 7/25} = \mp \frac{24}{7}$.Given the options,the magnitude is $\frac{24}{7}$.

If $\sin x + \cos x = \frac{1}{5},$ then $\tan 2x$ is

Similar Questions

If $\sin 21^{\circ} = \frac{x}{y},$ then $\sec 21^{\circ} - \sin 69^{\circ}$ is equal to

$\tan \frac{2\pi}{5} - \tan \frac{\pi}{15} - \sqrt{3} \tan \frac{2\pi}{5} \tan \frac{\pi}{15}$ is equal to

$ABC$ is a triangle. If $\sin \left(\frac{A+B}{2}\right) = \frac{\sqrt{3}}{2}$,then the value of $\sin \frac{C}{2}$ is

If $7 \sin ^{2} \theta+3 \cos ^{2} \theta=4,$ then $\tan \theta=$

If $\tan A$ and $\tan B$ are the roots of the quadratic equation $3x^2 - 10x - 25 = 0$,then the value of $3 \sin^2 (A + B) - 10 \sin (A + B) \cos (A + B) - 25 \cos^2 (A + B)$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module