(cosecA-sinA).(secA-cosA)=1/tanA+cotA
Asked by
| 29th Aug, 2012,
09:14: PM
Expert Answer:
LHS
= (cosecA sinA).(secA cosA)
= (1/sinA - sinA).(1/cosA - c0sA)
= [(1-Sin^2A)/SinA].[[1-cos^2A)/CosA]
= [cos^2A/SinA].[Sin^2A/CosA]
= cosA.SinA
LHS
= 1/(tanA + cotA)
= 1/((SinA/CosA + CosA/SinA)
= cosA.SinA / (sin^2A + cos^2A)
= cosA.SinA {since sin^2A + cos^2A = 1}
So
RHS = LHS
= (cosecA sinA).(secA cosA)
= (1/sinA - sinA).(1/cosA - c0sA)
= [(1-Sin^2A)/SinA].[[1-cos^2A)/CosA]
= [cos^2A/SinA].[Sin^2A/CosA]
= cosA.SinA
LHS
= 1/(tanA + cotA)
= 1/((SinA/CosA + CosA/SinA)
= cosA.SinA / (sin^2A + cos^2A)
= cosA.SinA {since sin^2A + cos^2A = 1}
So
RHS = LHS
Answered by
| 29th Aug, 2012,
10:38: PM
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