Math, asked by anupriyaambika54461, 9 months ago

Cosec theta + cot theta = m , proof it m square - 1 / m square + 1 = cos A

Answers

Answered by sandy1816

Step-by-step explanation:

cosecA+cotA=m

LHS,

m²-1/m²+1

=(cosecA+cotA)²-1/(cosecA+cotA)²+1

=(cosec²A+cot²A+2cosecAcotA-1)/(cosec²A+cot²A+2cosecAcotA+1)

=(cot²A+cot²A+2cosecAcotA)/(cosec²A+cosec²A+2cosecAcotA)

=(2cot²A+2cosecAcotA)/(cosec²A+2cosecAcotA)

=2cotA(cotA+cosecA)/2cosecA(cosecA+cotA)

=cotA/cosecA

=cosA

Answered by umiko28

$<body \: bgcolor = "'red "> \huge\red{ \mathbb{QUESTION}} \\ \\ </p><p>Cosec \theta + cot \theta = m , proof \: it \: m^{2} - 1 / m ^{2}+ 1 = cos A \ \\ \\ \huge\purple{ \mathbb{SOLUTION \to: }} \\ \\ here, \: cosecA+cotA=m\\ \\ \bf\ now,LHS, \\ \\ {m}^{2} -1/ {m}^{2} +1 \\ \\ \bf\ \implies:{(cosecA+cotA)}^{2}-1/{(cosecA+cotA)}^{2}+1 \\ \\ \bf\ \implies({cosec}^{2} A+{cot}^{2} A+2cosecAcotA-1)/({cosecA}^{2} +cotA+2cosecAcotA+1) \\ \\ \bf\ \implies: ({cot}^{2} A+{cot}^{2} A+2cosecAcotA)/({cosec}^{2} A+{cosec}^{2} A+2cosecAcotA)\\ \\ \bf\ \implies:({2cot}^{2} A+2cosecAcotA)/({cosec}^{2} A+2cosecAcotA) \\ \\ \bf\ \implies:2cotA(cotA+cosecA)/2cosecA(cosecA+cotA) \\ \\ \bf\ \implies: cotA/cosecA\\ \\ \bf\ \implies:cosA$

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