I have tried to substitute $\tan (x)$ and $\cot (x)$ with sine and cosine and it got really messy

$$1-\frac{\sin^2x}{1+\cot(x)}-\frac{\cos^2x}{1+\tan x}=?$$

a) $1$

b) $\sin^2x$

c) $\sin (x)+\cos (x)$

d) $\sin (x)\cdot \cos (x)$


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  • $\begingroup$ Have you at least calculated some values so you know which of $(a, b, c, d)$ to prove? $\endgroup$ – orlp Apr 19 '17 at 18:55

We can write this as : here I put $$\cot x= \frac{\cos x}{\sin x}$$ and $$\tan x= \frac{\sin x}{\cos x}$$

$$\sin^2x+\cos^2x-\frac{\sin^3x}{\sin x+\cos x}-\frac{\cos^3x}{\sin x+\cos x}$$

Now take take the $\operatorname{lcm}$ and you'll get something like this:

$$\frac {\sin^3x+\cos^3x+ \sin x\cos^2x+\cos x \sin^2x-\sin^3x-\cos^3x}{\sin x+\cos x}$$ then $$\frac{\sin x\cos x(\sin x+\cos x)}{\sin x +\cos x}$$ hence final asnwer is

d) $\sin x.\cos x$

  • $\begingroup$ ItiShree: Tip for formatting your posts. When it comes to trig functions, you can properly format the function using, for example `\cos x , \sin x, \tan x' etc. (simply using a backward slash immediately prior to the trig function you'd like to use.) $\endgroup$ – Namaste Apr 19 '17 at 19:30
  • $\begingroup$ It is me who should thank you kind person for you detailed explanation $\endgroup$ – L.B Apr 19 '17 at 19:30
  • $\begingroup$ @amWhy Thanks , I'm new so working on it. $\endgroup$ – Iti Shree Apr 19 '17 at 19:31
  • $\begingroup$ @L.B No need to thank, it's my pleasure helping you. $\endgroup$ – Iti Shree Apr 19 '17 at 19:31
  • $\begingroup$ @amWhy i appreciate it, thank you $\endgroup$ – L.B Apr 19 '17 at 19:32

note that $$\cot(x)=\frac{\cos(x)}{\sin(x)}$$ and $$\tan(x)=\frac{\sin(x)}{\cos(x)}$$ the result is $$\sin(x)\cos(x)$$ we have $$\frac{(1+\cot(x))(1+\tan(x))-\sin^2(x)(1+\tan(x))-\cos^2(x)(1+\cot(x))}{(1+\cot(x))(1+\tan(x))}$$ multiplying out and we otain $$\frac{1+\frac{\cos(x)}{\sin(x)}+\frac{\sin(x)}{\cos(x)}-\frac{\sin^3(x)}{\cos(x)}-\frac{\cos(x)^3}{\sin(x)}}{(1+\cot(x))(1+\tan(x))}$$ and this is $$\frac{1+\frac{\cos(x)(1-\cos^2(x))}{\sin(x)}+\frac{\sin(x)(1-\sin^2(x))}{\cos(x)}}{(1+\cot(x))(1+\tan(x))}$$ we will work further: $$\frac{1+2\sin(x)\cos(x)}{2+\frac{\cos(x)}{\sin(x)}+\frac{\sin(x)}{\cos(x)}}$$ and this is $$\frac{(1+2\sin(x)\cos(x))\sin(x)\cos(x)}{2\sin(x)\cos(x)+1}=\sin(x)\cos(x)$$

  • 2
    $\begingroup$ why the $-2$, what is wrong with my result? $\endgroup$ – Dr. Sonnhard Graubner Apr 19 '17 at 18:56
  • $\begingroup$ I have no issues with your result, I have issues with your argument. In it's current state you are missing steps to show how you went from substituting $\cot$/$\tan$ for their quotients to the final result. $\endgroup$ – orlp Apr 19 '17 at 18:58
  • $\begingroup$ Could you please explain in more detail how you came to this result because tan(x)=sine/cosine and cotx= cos/sin i know that already and tried that, there is something i missed and i would like to know $\endgroup$ – L.B Apr 19 '17 at 18:58
  • 1
    $\begingroup$ ok i will insert steps,ok? $\endgroup$ – Dr. Sonnhard Graubner Apr 19 '17 at 18:59
  • $\begingroup$ That would be great! $\endgroup$ – L.B Apr 19 '17 at 19:00

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