Question

help attachment

Answer 1

Q1 plese

Answer 2

tan(pi-x)= -tanx =-k
tan(pi/2-x) = cotx =1/k
sinx= tanx.secx = k/(sqrt(1+k*k)

Answer 3

can u elaborate the second one please why cotx?

Answer 4

in a r8 angled trngle (for simplicity) the complemnt of x is pi/2 -x thus the opposite side and the adjacent sides get interchanged

Answer 5

can u illustrate it graphically

Answer 6

um lost

Answer 7

otherwise use compound angle formulas for trigos

Answer 8

for the first one i know that tan(-thita)=-tan(thita) ican apply this formula to the first question but the second question i cant

Answer 9

\[\tan [(1/2)\pi-x] = (tanx)^{-1} = k^{-1} = 1/k\]