If sec x + tan x = 3, x in (0, frac{pi}{2}),t | vedclass

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

Question

(B) Given $\sec x + 	an x = 3$  $(1)$We know that $\sec^2 x - 	an^2 x = 1$Using the identity $a^2 - b^2 = (a - b)(a + b)$,we get $(\sec x - 	an x)(

Vedclass · Accepted Answer

$\frac{4}{5}$

Vedclass · Answer

(B) Given $\sec x + 	an x = 3$  $(1)$We know that $\sec^2 x - 	an^2 x = 1$Using the identity $a^2 - b^2 = (a - b)(a + b)$,we get $(\sec x - 	an x)(\sec x + 	an x) = 1$Substituting $(1)$,we have $\sec x - 	an x = \frac{1}{3}$  $(2)$Adding $(1)$ and $(2)$,we get $2 \sec x = 3 + \frac{1}{3} = \frac{10}{3}$Thus,$\sec x = \frac{5}{3}$,which implies $\cos x = \frac{3}{5}$Since $x \in (0, \frac{\pi}{2})$,$\sin x$ is positive.$\sin x = \sqrt{1 - \cos^2 x} = \sqrt{1 - (\frac{3}{5})^2} = \sqrt{1 - \frac{9}{25}} = \sqrt{\frac{16}{25}} = \frac{4}{5}$

If $\sec x + \tan x = 3, x \in (0, \frac{\pi}{2})$,then $\sin x =$

Similar Questions

If $\sin (A + B) = 1$ and $\cos (A - B) = \frac{\sqrt{3}}{2}$,then the smallest positive values of $A$ and $B$ are

Find the radian measure corresponding to the following degree measure: $520^{\circ}$

If $\sqrt{3} \cos \theta + \sin \theta > 0$,then:

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

If $\tan \theta = \frac{\sin \alpha - \cos \alpha}{\sin \alpha + \cos \alpha}$,then $\sin \alpha + \cos \alpha$ and $\sin \alpha - \cos \alpha$ are equal to:

Vedclass Test Series

Exam Paper Generator

Online Exam Module