Question

The graphs of $y=\sin x,y=\cos x,y=\tan x$ & $y=\csc x$ are drawn on the same axes from $0$ to $\pi /2$. A vertical line is drawn through the points where the graphs of $y=\cos x$ & $y=\tan x$ cross, intersecting the other two graphs at the points $A$ & $B$. The length of the line segment $AB$ is ?

• A. $1$
• B. $\frac{\sqrt{5}-1}{2}$
• C. $\sqrt{2}$
• D. $\frac{\sqrt{5}+1}{2}$

Answer 1

The correct option is A $1$

$y=cosxandy=tanxintersect$

$cosx=tanx$

$cosx=\frac{sinx}{cosx}$

${cos}^{2}x=sinx$

$1-{sin}^{2}x=sinx$

${sin}^{2}x+sinx-1=0$

$sinx=\frac{-1\pm \sqrt{1^2-4\times 1(-1)}}{2\times 1}$

$sinx=\frac{-1\pm \sqrt{5}}{2}$

$sinx=\frac{-1+\sqrt{5}}{2}$

$equationoflinepassingthroughthepointy=cosxandy=tanxisx=h$

$sinh=\frac{-1+\sqrt{5}}{2}=BC$

$cosech=\frac{1}{sinh}=\frac{2}{-1+\sqrt{5}}=AC$

$AB=AC-BC=\frac{2}{-1+\sqrt{5}}-\frac{-1+\sqrt{5}}{2}$

$\frac{4-(-1+{\sqrt{5}}^2)}{2(-1+\sqrt{5})}$

$\frac{4-(1+5-2\sqrt{5})}{2(-1+\sqrt{5})}$

$\frac{4-6(+2\sqrt{5})}{2(-1+\sqrt{5})}$

$\frac{-2+2\sqrt{5}}{-2+2\sqrt{5}}$