If A+B+C=90o, then cotAcotBcotCcotA+cotB+cotC is equal to:

A+B=90o−CTake cot on both sides, we getcot(A+B)=cot(90o−C)⇒cotA+cotBcotAcotB−1=tanC=cotC1Cross-Multiply both sides, we getcotAcotBcotC−cotC=cotA+cotB⇒cotAcotBcotB=cotA+cotB+cotC⇒cotAcotBcotCcotC+cotB+cotC=1

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If A+B+C=90^o, then { A+ B+ C}{ A B C} is equal to:

Question

If $A + B + C = 9 0^{o}$ , then $\frac{cot A + cot B + cot C}{cot A cot B cot C}$ is equal to:

1

cot A cos B cot C

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Lucas Discussed

A + B + C = 9 0^{o}

, then

\frac{cot A + cot B + cot C}{cot A cot B cot C}

is equal to:

9 mins ago

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Text solutionVerified

A + B = 9 0^{o} - C

Take

cot

on both sides, we get

cot (A + B) = cot (9 0^{o} - C)

\Rightarrow \frac{cot A cot B - 1}{cot A + cot B} = tan C = \frac{1}{cot C}

Cross-Multiply both sides, we get

cot A cot B cot C - cot C = cot A + cot B

\Rightarrow cot A cot B cot B = cot A + cot B + cot C

\Rightarrow \frac{cot C + cot B + cot C}{cot A cot B cot C} = 1

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Question Text	If $A + B + C = 9 0^{o}$ , then $\frac{cot A + cot B + cot C}{cot A cot B cot C}$ is equal to:
Answer Type	Text solution:1
Upvotes	150

If A+B+C=90o, then cotAcotBcotCcotA+cotB+cotC​ is equal to:

Text solutionVerified

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If $A + B + C = 9 0^{o}$ , then $\frac{cot A + cot B + cot C}{cot A cot B cot C}$ is equal to: