# I have this trigonometric expression to prove [closed]

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

## closed as off-topic by Namaste, Jean-Claude Arbaut, Henrik, Chris Godsil, ArnaldoApr 20 '17 at 14:01

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Namaste, Jean-Claude Arbaut, Henrik, Chris Godsil, Arnaldo
If this question can be reworded to fit the rules in the help center, please edit the question.

• Have you at least calculated some values so you know which of $(a, b, c, d)$ to prove? – 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) $\sin x.\cos x$

• 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.) – Namaste Apr 19 '17 at 19:30
• It is me who should thank you kind person for you detailed explanation – L.B Apr 19 '17 at 19:30
• @amWhy Thanks , I'm new so working on it. – Iti Shree Apr 19 '17 at 19:31
• @L.B No need to thank, it's my pleasure helping you. – Iti Shree Apr 19 '17 at 19:31
• @amWhy i appreciate it, thank you – 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 $$\frac{(1+2\sin(x)\cos(x))\sin(x)\cos(x)}{2\sin(x)\cos(x)+1}=\sin(x)\cos(x)$$

• why the $-2$, what is wrong with my result? – Dr. Sonnhard Graubner Apr 19 '17 at 18:56
• 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. – orlp Apr 19 '17 at 18:58
• 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 – L.B Apr 19 '17 at 18:58
• ok i will insert steps,ok? – Dr. Sonnhard Graubner Apr 19 '17 at 18:59
• That would be great! – L.B Apr 19 '17 at 19:00