How do you prove that tan(180-A)sin(270+A)cosec(360-A)=-1?

1 Answer
Jun 27, 2018

Please see below.

Explanation:

Here,

tan(180^circ-A)=-tanA to[becauseII^(nd)Quadrant , tantheta <0]

sin(270^circ+A)=-cosAto[becauseIV^(th)Quadrant,sintheta <0]

csc(360^circ-A)=-cscAto[becauseIV^(th)Quadrant,csctheta <0]

So,

LHS=tan(180^circ-A)+sin(270^circ+A)+csc(360^circ-A)

LHS=(-tanA)(-cosA)(-cscA)

LHS=- sinA/cancelcosAxxcancelcosAxxcscAto[becausetantheta=sintheta/costheta]

LHS=-sinAxx1/sinAto[becausecsctheta=1/sintheta]

LHS=-1

Note: We have taken

180to180^circ,270to270^circ and 360to360^circ