# cot(x-y)= cotxcoty+1/coty-cotx?

I need help with solving this!! I have to verify the identity. Thank you!

### 7 Answers

- gudspelingLv 71 decade agoFavorite Answer
LHS

= cot(x-y)

= 1/tan(x-y)

= 1/((tan(x) - tan(y))/(1 + tan(x)tan(y)))

= (1 + tan(x)tan(y)) / (tan(x) - tan(y)) ------- now divide the numerator and denominator by tan(x)tan(y)

= (cot(x)cot(y) + 1) / (cot(y) - cot(x))

= RHS

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- 4 years ago
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RE:

cot(x-y)= cotxcoty+1/coty-cotx?

I need help with solving this!! I have to verify the identity. Thank you!

Source(s): cotxcoty 1 coty cotx: https://biturl.im/xLhso- Login to reply the answers

- 1 decade ago
cot(x-y)= cos(x-y)/sin(x-y)=cosx.cosy+sinx.siny/sinx.cosy-cosx.siny

{divide Nr&Dr by sinx.siny}

=cotxcoty+1/coty-cotx.=RHS

Hence proved.

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- KristenLv 44 years ago
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cot(x+y) = (cotxcoty-1)/(cotx+coty)? Girl, please read patiently. It is NOT tough but needs patience. LHS = cot(x+y) = cos(x+y) / sin(x+y) = [(cosx*cosy - sinx*siny) / (sinx*cosy + cosx*siny)] Divide numerator and denominator by sinx*siny, we get (cotxcoty-1)/(coty+cotx), = (cotxcoty-1)/(cotx+coty) = RHS

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- 1 decade ago
cot(x-y)= cos(x-y)/sin(x-y)

or = [cosxcosy +sinxsiny]/[sinxcosy-cosxsiny]

= cotxcoty+1/coty - cotx

[dividing numeretor and denomerator by sinxsiny]

now further clarification to prove sin(x-y) and cos(x+y) etc

one example shown below .others can be found in your text book

To prove cos(x ± y) = cosx cosy ∓ sinx siny

then

sin(x + y) = cos(π/2 - (x+y)) = cos((π/2 - x) - y)

= cos(π/2 - x) cosy + sin(π/2 - x) siny

= sinx cosy + cosx siny

now, to prove that cos(x + y) = cosx cosy - sinx siny .. . .. .

this used the distance formula:

let P(1,0), A(cosx, sinx) , B(cosy, siny) , C(cos(x-y), sin(x-y))

the angle from the positive x-axis to A is x.

the angle from the positive y-axis to B is y.

thus the angle between them is x - y , say, x > y.

thus the distance bet AB is the same distance as CP

then

d(AB) = √[(cosx - cosy)^2 + (sinx - siny)^2]

= √[cos^2x - 2cosx cosy + cos^2y + sin^2x - 2sinxsiny + sin^2y]

= √[2 - 2(cosx cosy + sinx siny)]

meanwhile

d(CP) = √[(1- cos(x-y)^2 + sin^2(x-y)]

= √[1 - 2cos(x-y) + cos^2(x-y) + sin^2(x-y)]

= √[2 - 2cos(x-y)]

thus

cos(x - y) = (cosx cosy + sinx siny)

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- 1 decade ago
LHS

= cot(x-y)

= 1/tan(x-y)

= 1/((tan(x) - tan(y))/(1 + tan(x)tan(y))) {(formula of tan(x-y)}

= (1 + tan(x)tan(y)) / (tan(x) - tan(y)) ------- now divide the numerator and denominator by tan(x)tan(y)

= (cot(x)cot(y) + 1) / (cot(y) - cot(x))

= RHS

It can use as formula also

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