Question

A+B+C=90 ; cotA+cotB+cotC= cotA.cotB.cotC​

Answer 1

Step-by-step explanation:

1. Taking LHS
2. CotA+CotB+CotC
3. Taking Cot common Cot (A+B+C)
4. Cot90 =0
5. CotA. CotB. CotC =0

Answer 2

ANSWER:

Given:

• A+B+C=90

To Prove:

• cotA+cotB+cotC= cotA•cotB•cotC

Proof:

$\text{We are given that,}\\\\:\longrightarrow A+B+C=90\\\\:\implies A+B=90-C\\\\\text{Introducing cot function both sides,}\\\\:\implies\cot(A+B)=\cot(90-C)\\\\\text{We know that:}\\\\:\hookrightarrow\cot(X+Y)=\dfrac{\cot(X)\cot(Y)-1}{\cot(Y)+\cot(X)}\:\&\:\cot(90-\theta)=\tan(\theta)\\\\\text{So,}\\\\:\implies\dfrac{\cot(A)\cot(B)-1}{\cot(B)+\cot(A)}=\tan(C)\\\\\text{We know that,}\\\\:\hookrightarrow\tan(\theta)=\dfrac{1}{\cot(\theta)}\\\\\text{So,}\\\\:\implies\dfrac{\cot(A)\cot(B)-1}{\cot(B)+\cot(A)}=\dfrac{1}{\cot(C)}\\\\\text{On cross-multiplying,}\\\\:\implies[\cot(A)\cot(B)-1]\times[\cot(C)]=\cot(A)+\cot(B)\\\\:\implies\cot(A)\cot(B)\cot(C)-\cot(C)=\cot(A)+\cot(B)\\\\\text{Transposing cotC to RHS,}\\\\:\implies\cot(A)\cot(B)\cot(C)=\cot(A)+\cot(B)+\cot(C)\\\\\bf{:\implies\cot(A)+\cot(B)+\cot(C)=\cot(A)\cot(B)\cot(C)}\\\\\text{\bf{Hence proved!!!}}$

Formulae Used:

• cot(X+Y)=(cotX×cotY-1)/(cotY+cotX)
• cot(90-X)=tanX
• tanX=1/(cotX)