• Aug 26th 2006, 06:32 PM
Prove the identities:
1.) cot (A + B) = (cotAcotB - 1) / (cotB + cotA)
2.) cot (A - B) = (cotAcotB + 1) / (cotB - cotA)
3.) sec (A + B) = (secAsecB) / (1 - tanAtanB)
4.) csc (A + B) = (cscAcscB) / (cotB + cotA)
• Aug 26th 2006, 06:52 PM
ThePerfectHacker
Quote:

Prove the identities:
1.) cot (A + B) = (cotAcotB - 1) / (cotB + cotA)

$\frac{\cos (x+y)}{\sin (x+y)}$
$\frac{\cos x\cos y-\sin x\sin y}{\sin x\cos y+\sin y\cos x}$
Divide numerator and denominator by $\sin x\sin y$, thus,
$\frac{\cot x\cot y-1}{\cot y+\cot x}$
• Aug 26th 2006, 06:56 PM
ThePerfectHacker
Quote:

2.) cot (A - B)

Simple,
$\cot (x-y)=\cot (x+(-y))$
$\frac{\cot x\cot (-y)-1}{\cot x+\cot (-y)}$
$\frac{-\cot x\cot y-1}{\cot x-\cot y}$
Multiply numerator and denominator by (-1):
$\frac{\cot x\cot y+1}{\cot y-\cot x}$
• Aug 26th 2006, 06:59 PM
ThePerfectHacker
Quote:

3.) sec (A + B) = (secAsecB) / (1 - tanAtanB)

$\frac{1}{\cos (x+y)}=\frac{1}{\cos x\cos y-\sin x\sin y}$
Divide numerator and denominator by $\cos x\cos y$
Thus,
$\frac{\frac{1}{\cos x\cos y}}{1-\tan x\tan y}$

$\frac{\sec x\sec y}{1-\tan x\tan y}$
• Aug 26th 2006, 07:02 PM
ThePerfectHacker
Quote:

4.) csc (A + B) = (cscAcscB) / (cotB + cotA)

$\frac{1}{\sin (x+y)}=\frac{1}{\sin x\cos y+\cos x\sin y}$
Divide numerator and denominator by $\sin x\sin y$
Thus,
$\frac{\frac{1}{\sin x\sin y}}{\cot y+\cot x}$

$\frac{\csc x\csc y}{\cot x+\cot y}$
• Aug 27th 2006, 01:54 AM
Quote:

Originally Posted by ThePerfectHacker
Simple,
$\cot (x-y)=\cot (x-(-y))$
$\frac{\cot x\cot (-y)-1}{\cot x+\cot (-y)}$
$\frac{-\cot x\cot y-1}{\cot x-\cot y}$
Multiply numerator and denominator by (-1):
$\frac{\cot x\cot y+1}{\cot y-\cot x}$

got a q sir
how come cot (x-y) became equal to cot (x-(-y)) ?
• Aug 27th 2006, 02:00 AM
Glaysher
Quote: