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)

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,
$\frac{\cot x\cot y-1}{\cot y+\cot x}$

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):
$\frac{\cot x\cot y+1}{\cot y-\cot x}$

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,
$\frac{\frac{1}{\cos x\cos y}}{1-\tan x\tan y}$

$\frac{\sec x\sec y}{1-\tan x\tan y}$

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,
$\frac{\frac{1}{\sin x\sin y}}{\cot y+\cot x}$

$\frac{\csc x\csc y}{\cot x+\cot y}$

6. 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)) ?

got a q sir
how come cot (x-y) became equal to cot (x-(-y)) ?
It didn't. Its a typo. It should read cot(x - y) = cot(x + (-y))

Then he uses the answer from part (a)

8. Problem Fixed, thank you for noticing.

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# cot (a b)=cot a cotb-1/

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