if cosecA +cotA = m then show that m^2-1/m^2+1 = cos A
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Answered by
156
cosec A + cot A =m-----(1)
lhs = (m²-1)/(m²+1)
= [(cosec A+cot A)² - 1]/[(cosec A +cot)²+1] from (1)
= [ cosec² A +2cosecAcotA+cot² A-1]/[cosec²A +2cosec AcotA+cot ²A +1]
= [(cosec ²A-1 )+2cosec AcotA+cot²A]/[cosec²A +2cosecAcotA +(cot²A+1)]
= [cot²A+2cosecAcotA+cot²A]/[cosec²A+2cosecAcotA+cosec²A]
{since cosec²A-1 = cot²A and cot²A +1 = cosec²A]
=[2cot²A+2cosecAcotA]/[2cosec²A+2cosecAcotA]
=[2cotA(cotA+cosecA)]/ [2cosecA(cosecA+cotA)]
after cancellation
=cotA/cosecA
=(cosA/sinA)/(1/sinA)
= cosA
=rhs
lhs = (m²-1)/(m²+1)
= [(cosec A+cot A)² - 1]/[(cosec A +cot)²+1] from (1)
= [ cosec² A +2cosecAcotA+cot² A-1]/[cosec²A +2cosec AcotA+cot ²A +1]
= [(cosec ²A-1 )+2cosec AcotA+cot²A]/[cosec²A +2cosecAcotA +(cot²A+1)]
= [cot²A+2cosecAcotA+cot²A]/[cosec²A+2cosecAcotA+cosec²A]
{since cosec²A-1 = cot²A and cot²A +1 = cosec²A]
=[2cot²A+2cosecAcotA]/[2cosec²A+2cosecAcotA]
=[2cotA(cotA+cosecA)]/ [2cosecA(cosecA+cotA)]
after cancellation
=cotA/cosecA
=(cosA/sinA)/(1/sinA)
= cosA
=rhs
Answered by
35
Answer:
Step-by-step explanation:cosec A + cot A =m-----(1)
lhs = (m²-1)/(m²+1)
= [(cosec A+cot A)² - 1]/[(cosec A +cot)²+1] from (1)
= [ cosec² A +2cosecAcotA+cot² A-1]/[cosec²A +2cosec AcotA+cot ²A +1]
= [(cosec ²A-1 )+2cosec AcotA+cot²A]/[cosec²A +2cosecAcotA +(cot²A+1)]
= [cot²A+2cosecAcotA+cot²A]/[cosec²A+2cosecAcotA+cosec²A]
{since cosec²A-1 = cot²A and cot²A +1 = cosec²A]
=[2cot²A+2cosecAcotA]/[2cosec²A+2cosecAcotA]
=[2cotA(cotA+cosecA)]/ [2cosecA(cosecA+cotA)]
after cancellation
=cotA/cosecA
=(cosA/sinA)/(1/sinA)
= cosA
=rhs
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