The value of Cosec A-l is
Answers
Answer:
Does Cosec A - 1 / cosec A + 1 = (cos A / 1+ sin A) square?
(Cosec A - 1)/(Cosec A + 1)
= ((1/Sin A) - 1)/((1/Sin A) + 1))
= (1 - Sin A)/(1 + Sin A)
divide each term by (1+Sin A)
= (1 - Sin A^2)/(1 + Sin A)^2
= (Cos A/(1 + Sin A))^2
LHS : (cosecA -1)/(cosecA + 1) = [(1/sinA) - 1] / [ (1/sinA) + 1]
= [(1-sinA)/(1 + sinA)]
Multiplying both numerator and denominator with (1 + sinA)
= [(1-sinA)*(1+sinA)] / [(1+sinA)*(1 + sinA)]
= [1-square(sinA)][square(1 +sinA)]
= square(cosA)/square(1 + sinA)
= square[cosA / (1 + sinA)]
= RHS.
cosec A = 1/sin A
(cosec A - 1)/(cosec A + 1) = (1 - sin A)/(1 + sin A) -----------> 1
multiply numerator and denominator in 1 by (1 + sin A) to get
(1 - sin A)/(1 + sin A) * (1 + sin A)/(1 + sin A) =
(1 - sin^2 A)/1 + sin^2 A) --------------> 2
now 1 - sin^2 A = cos^2 A --------------> standard identity cos^2 A + sin^2 A = 1
so we get from 2 the expression simplified as cos^2 A/(1 + sin A)^2 =
(cos A/(1 + sin A))^2
Is cosec A + 1/ cosec A - 1 equal to (secA+tanA) ^2?
How do I prove that (cot A + cosec A -1) / (cotA - cosecA +1) = (1+cosA) / sinA?
How do I prove cot A+cosec A -1/cot A -cosec A+1 =1+cos A/sin A?
(cosecA-1)/(cosecA+1)=[cosA/(1+sinA)]^2
LHS Multiply above and below bysinA.
=(1-sinA)/(1+sinA) , Again multiply above and below by (1+sinA).
=(1-sin^2A)/(1+sinA)^2
=cos^2A/(1+sinA)^2
=[cosA/(1+sinA)]^2 , proved
Yes,if we take LHS and RHS and evaluate them individually with the help of formula ,then we Will get 1-sin A/1+sin A on both the LHS and the RHS
Yes.its solution is as follows if u divide and multiply by coesc+1 then you gate ;(cosec^2a–1)/(coseca+1)^2=cot^2a/(coseca+1)^2=The given answer
Is cosec A + 1/ cosec A - 1 equal to (secA+tanA) ^2?
How do I prove that (cot A + cosec A -1) / (cotA - cosecA +1) = (1+cosA) / sinA?
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