Prove that
1-sin^2X/1+cotX - cos^2X/1+tanX = sinXcosX
Much thanks in advance.
Hello, ahling!
We have: .$\displaystyle 1 - \frac{\sin^2\!x}{1 + \dfrac{\cos x}{\sin x}} - \frac{\cos^2\!x}{1 + \dfrac{\sin x}{\cos x}}$Prove that: .$\displaystyle 1 - \frac{\sin^2\!x}{1 + \cot x}- \frac{\cos^2\!x}{1 + \tan x} \:=\:\sin x\cos x$
Multiply the first fraction by $\displaystyle \frac{\sin x}{\sin x}$ , the second by $\displaystyle \frac{\cos x}{\cos x}$
. . $\displaystyle 1 - \frac{\sin^3\!x}{\sin x + \cos x} - \frac{\cos^3\!x}{\cos x + \sin x} \;\;=\;\;1 - \frac{sin^3\!x + \cos^3\!x}{\sin x + \cos x}$
. . $\displaystyle = \;\;1 - \frac{(\sin x + \cos x)(\sin^2\!x - \sin x\cos x + \cos^2\!x)}{\sin x + \cos x}$
. . $\displaystyle = \;\;1 - (\underbrace{\sin^2\!x + \cos^2\!x}_{\text{This is 1}} - \sin x\cos x) \;\;=\;\;1 - (1 - \sin x\cos x)$
. . $\displaystyle = \;\sin x\cos x$ . . . ta-DAA!
And LateX is working !!
I have a little trouble believing that, but let's go with it...
1-sin^2X/1+cotX - cos^2X/1+tanX = sinXcosX
The "/1" are meaningless in this form, so...
1-sin^2X+cotX - cos^2X+tanX = sinXcosX
1-sin^2X = cos^2X <== Identity to Substitute
cos^2X+cotX - cos^2X+tanX = sinXcosX
cotX+tanX = sinXcosX
Since cotX = (cosX)/(sinX) and tanX = (sinX)/(cosX)
And Since cos^2X + sin^2X = 1
1/(sinXcosX) = sinXcosX
I think you would have a hard time selling that as an identity.
Note: I would do this on an exam. It is technically correct, but may have points taken off for being a smart aleck. I would also rewrite it as it probably is intended and work it that way. This may avoid the smart aleck points. Don't put up with sloppy notation. Certainly don't write any yourself. Everyone needs reminders to be clear - particularly if they want to be understood generally.
LaTeX Check: $\displaystyle \cos^{2}(x) + \sin^{2}(x) = 1$