{Cot (A+B).Cot A + 1} / {Cot A - Cot (A+B)} = Cot B
N Navo New member Joined Apr 8, 2013 Messages 1 Apr 8, 2013 #1 {Cot (A+B).Cot A + 1} / {Cot A - Cot (A+B)} = Cot B
S soroban Elite Member Joined Jan 28, 2005 Messages 5,586 Apr 8, 2013 #2 Hello, Navo! \(\displaystyle \text{Prove: }\:\dfrac{\cot(A\!+\!B)\!\cdot\!\cot A + 1}{\cot A - \cot(A\!+\!B)} \:=\:\cot B\) Click to expand... We have: .\(\displaystyle \displaystyle \frac{\cot(A\!+\!B)\!\cdot\!\cot A +1}{\cot A - \cot(A\!+\!B)} \;=\; \frac{\frac{1}{\tan(A+B)}\!\cdot\!\frac{1}{ \tan A} + 1}{\frac{1}{\tan A} - \frac{1}{\tan(A+B)}} \) Multiply by \(\displaystyle \frac{\tan A\cdot\tan(A+B)}{\tan A\cdot\tan(A+B)}:\;\;\dfrac{1+\tan A\tan(A\!+\!B)}{\tan(A\!+\!B) - \tan A} \;=\;\dfrac{1}{\dfrac{\tan(A\!+\!B) - \tan A}{1+\tan A\tan(A\!+\!B)}} \) Note that the denominator is: .\(\displaystyle \tan\big[(A\!+\!B)-A\big] \;=\;\tan B\) Therefore, we have: .\(\displaystyle \dfrac{1}{\tan B} \:=\:\cot B\)
Hello, Navo! \(\displaystyle \text{Prove: }\:\dfrac{\cot(A\!+\!B)\!\cdot\!\cot A + 1}{\cot A - \cot(A\!+\!B)} \:=\:\cot B\) Click to expand... We have: .\(\displaystyle \displaystyle \frac{\cot(A\!+\!B)\!\cdot\!\cot A +1}{\cot A - \cot(A\!+\!B)} \;=\; \frac{\frac{1}{\tan(A+B)}\!\cdot\!\frac{1}{ \tan A} + 1}{\frac{1}{\tan A} - \frac{1}{\tan(A+B)}} \) Multiply by \(\displaystyle \frac{\tan A\cdot\tan(A+B)}{\tan A\cdot\tan(A+B)}:\;\;\dfrac{1+\tan A\tan(A\!+\!B)}{\tan(A\!+\!B) - \tan A} \;=\;\dfrac{1}{\dfrac{\tan(A\!+\!B) - \tan A}{1+\tan A\tan(A\!+\!B)}} \) Note that the denominator is: .\(\displaystyle \tan\big[(A\!+\!B)-A\big] \;=\;\tan B\) Therefore, we have: .\(\displaystyle \dfrac{1}{\tan B} \:=\:\cot B\)
D Deleted member 4993 Guest Apr 8, 2013 #3 Navo said: {Cot (A+B).Cot A + 1} / {Cot A - Cot (A+B)} = Cot B Click to expand... You can directly use the formula: \(\displaystyle cot(x-y) \ = \ \dfrac{cot(x) * cot(y) + 1}{cot(x) - cot(y)}\)
Navo said: {Cot (A+B).Cot A + 1} / {Cot A - Cot (A+B)} = Cot B Click to expand... You can directly use the formula: \(\displaystyle cot(x-y) \ = \ \dfrac{cot(x) * cot(y) + 1}{cot(x) - cot(y)}\)
