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I am currently trying to solve the following equations : $$ \tan(180° - A)\sin(270°+A)\csc(360°-A) = -1 $$

I followed conventional wisdom and started by only simplifying one side of the equation (in this case the left) down. I ended up with this :

$$=-\tan A \cdot -\sin A \cdot \frac{1}{-\sin A}$$ $$ = -\tan A$$

But I could not get past there. I have solved these type of equations before, but this one is stumping me!

3 Answers 3

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You need to remember that-

(1) sine and cosecant functions are positive in $[0°,90°]$ and $[90°,180°]$.

$\sin 135°= \sin (90°+45°)= \cos 45°$

$\sin 225°= \sin (270°-45°)= - \cos 45°$

(2) cosine and secent functions are positive in $[0°,90°]$ and $[270°,360°]$

(3) tangent and cotangent functions are positive in $[0,90°]$ and $[180°,270°]$

And -

(1) in any trig function, addition/subtraction by $180°$ or $360°$ does not lead to change in the function.

(2) in any trig function, addition/subtraction by $90°$ or $270°$ leads to change in the function. (sine becomes cosine, tangent becomes cotangent and secant becomes cosecant and vice versa).

$ \tan(180° - A)= -\tan A$

$ \sin(270°+A) = -\cos A $

$ \csc(360°-A) = -\csc A$

$ \tan(180° - A)\sin(270°+A)\csc(360°-A)=(-\tan A)(-\cos A)(-\csc A)$

$=-(\frac{\sin A}{\cos A})(\cos A)(\frac{1}{\sin A})=-1 $

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    great response, but i still do not understand one part. why does sin(270+A) become NEGATIVE cosA. I follow until there due to getting positive cosA. I do not understand where I'm going wrong. Would you mind explaining it? – 2017-02-17
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    @JohnHon A way to understand this is by thinking of A as an acute angle. By adding 270 degrees to A, you are shifting it to the fourth quadrant where (270+A) lies between 270 and 360. In this quadrant, sign of sine changes. So, the negative sign appears. Sine is only positive when angle is between 0 to 180. As for changing to cosA, just remember that increment by 180 or 360 wont change the function but increment by 90 or 270 will. Another method, by adding or subtracting 360 sin (270+A) = sin (270+A-360) = sin (A-90) = - sin (90-A) = - cos A Hope this helps. – 2017-02-17
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    https://image.slidesharecdn.com/trigonometricratiosandidentities-130406033612-phpapp02/95/trigonometric-ratios-and-identities-5-638.jpg?cb=1365219430 – 2017-02-17
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$${\sin (270^{\circ} + A)=-\cos A} $$

1)Remember if Trigonometric ratio is applied on $n\frac{\pi}{2} ± \theta $, then

$(ⅰ)$ When $n$ is even,there is no change in the Trigonometric ratio (sign may be + or -).

$(ⅱ)$ When $n$ is odd , the change in Trigonometric ratio ( sign may be + or -) is as indicated below $$\sin↔\cos$$

2) If $\alpha$ is $n\pi - \theta $ then $\sin \alpha$ changes to $(-1)^{n+1} \sin \theta$

$\cos \alpha$ changes to $(-1)^n cos \theta$ and $\tan \alpha $ changes to $-\tan\theta$

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    Everything is correct,just one wrong this is what I wrote in answer^ – 2017-02-17
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    i am confused as to how one obtains this though – 2017-02-17
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    This does not provide an answer to the question. To critique or request clarification from an author, leave a comment below their post. - [From Review](/review/low-quality-posts/763738) – 2017-02-17
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    @JohnHon check my edit – 2017-02-17
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    @TheGeekGreek I edited – 2017-02-17
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We have $\tan(180^\circ -A) = -\tan A, \sin(270^\circ + A) = -\cos A, \csc(360^\circ -A) = -\csc A$.

Thus, we have $$\tan(180^\circ -A)\sin(270^\circ + A)\csc(360^\circ -A)$$ $$ = -\tan A (-\cos A)(-\csc A)$$ $$=(-\frac{\sin A}{\cos A})(-\cos A)(-\csc A)$$$$ = (\sin A)(-\csc A) = -1$$

Hope it helps.

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    @JohnHon If it helps, see here: http://myhandbook.info/form_trigono0.html – 2017-02-17