Question

A+B+C=90 then prove that cotA+cotB+cotC=cotAcotBcotC

Answer 1

here's the answer of your question

Answer 2

A + B + C = 90 then cot A + cot B + cot C = cot A cot B cotC is proved

Solution:

Given that A + B + C = 90

We have to prove that: cot A + cot B + cot C = cot A cot B cotC

Let us take the given information and solve

A+B+ C = 90

Which can be written as A + B = 90 – C

Taking cos on both sides, we get

cot(A + B) = cot(90 - C)   ----- eqn 1

Using the trignometric identity:

$\begin{array}{l}{\cot (A+B)=\frac{\cot A \times \cot B-1}{\cot A+\cot B}} \\\\ {\text {And } \cot (90-C)=\tan C}\end{array}$

Applying these in eqn 1 we get,

$\begin{array}{l}{\frac{\cot A \times \cot B-1}{\cot A+\cot B}=\tan C} \\\\ {\frac{\cot A \times \cot B-1}{\cot A+\cot B}=\frac{1}{\cot C}}\end{array}$

On cross-multiplication we get

$\cot C [\cot A \times \cot B-1]=\cot A+\cot B$

$\cot C \times \cot A \times \cot B-\cot C=\cot A+\cot B$

$\cot A \times \cot B \times \cot C=\cot A+\cot B+\cot C$

Thus proved

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