sinA/1+cosA= cosecA-cotA
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Answered by
1
l.h.s.
sinA /1+ cos A
sinA(1-cosA)/(1+cosA)(1-cosA)
in the denominator you will get
1-cos^2A which is equal to sin^2A
therefore sinA will be canceled from the numerator
and you will be left with
1-cosA/sinA
after separating them you will get your answer that is cosecA - cotA
sinA /1+ cos A
sinA(1-cosA)/(1+cosA)(1-cosA)
in the denominator you will get
1-cos^2A which is equal to sin^2A
therefore sinA will be canceled from the numerator
and you will be left with
1-cosA/sinA
after separating them you will get your answer that is cosecA - cotA
sriram77:
ur welcome
Answered by
0
LHS
sinA/(1+cosA)
Dividing Both numerator and denominator by (sinA)
we get,
1/(1/SinA+cosA/sinA)
=1/(cosecA+cotA)
(since, cosA/sinA=cotA)
Now by rationalising,
we get...
cosecA-cotA/{(cosecA)²-(cotA)²)}
cosecA-cotA/{(cosec²A)-(cot²A)}
=cosecA-cosA = RHS
(since, Cosec²A-Cot²A=1)
sinA/(1+cosA)
Dividing Both numerator and denominator by (sinA)
we get,
1/(1/SinA+cosA/sinA)
=1/(cosecA+cotA)
(since, cosA/sinA=cotA)
Now by rationalising,
we get...
cosecA-cotA/{(cosecA)²-(cotA)²)}
cosecA-cotA/{(cosec²A)-(cot²A)}
=cosecA-cosA = RHS
(since, Cosec²A-Cot²A=1)
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