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

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Answer 1 · 2018-06-27T12:03:43

Please see below.

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#