Rezwanwan95
25

# If tan x = 3/4 and 0◦ ≤ x◦ ≤ 90◦, then cos x =? Please explain the work behind the solution

$tanx=\frac{3}{4}\\\\tanx=\frac{sinx}{cosx}\\\\\frac{sinx}{cosx}=\frac{3}{4}\to sinx=\frac{3}{4}cosx\ (*)\\\\\\sin^2x+cos^2x=1\\\\substitute\ (*)\\\\\left(\frac{3}{4}cosx\right)^2+cos^2x=1\\\\\frac{9}{16}cos^2x+cos^2x=1$ $\frac{15}{16}cos^2x=1\ /\cdot\frac{16}{25}\\\\cos^2x=\frac{16}{25}\\\\cosx=\sqrt\frac{16}{25}\\\\cosx=\frac{4}{5}$