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

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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

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

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