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CosA/1-tanA+sinA/1-cotA
CosA/1-sinA/cosA+sinA/1-cosA/sinA
CosA/cosA-sinA/cosA+sinA/sinA-cosA/sinA
Cos^2A/cosA-sinA-sin^2A/sinA-cosA
Cos^2A-sin^2A/cosA-sinA
Cancel it you will get
CosA+sinA
CosA/1-sinA/cosA+sinA/1-cosA/sinA
CosA/cosA-sinA/cosA+sinA/sinA-cosA/sinA
Cos^2A/cosA-sinA-sin^2A/sinA-cosA
Cos^2A-sin^2A/cosA-sinA
Cancel it you will get
CosA+sinA
Answered by
1
Answer:
cosA/ (1-tanA) - sinA/(1-cotA) = cosA + sinA
We know that:
tanA = sinA/cosA
cotA = cosA/sinA
Now substitute in L.H.S:
==> cosA/(1-sinA/cosA) - sinA/(1-cosA/sinA)
= cosA/[(cosA-sinA)/cosA] - sinA/[(sinA-cosA)/sinA]
= (cos^2 A - sin^2 A)/ (cosA-sinA)
= (cosA-sinA)(cosA+ sinA)/(cosA-sinA)
= cosA + sinA = R.H.S
PlZ mark as brainliest.
cosA/ (1-tanA) - sinA/(1-cotA) = cosA + sinA
We know that:
tanA = sinA/cosA
cotA = cosA/sinA
Now substitute in L.H.S:
==> cosA/(1-sinA/cosA) - sinA/(1-cosA/sinA)
= cosA/[(cosA-sinA)/cosA] - sinA/[(sinA-cosA)/sinA]
= (cos^2 A - sin^2 A)/ (cosA-sinA)
= (cosA-sinA)(cosA+ sinA)/(cosA-sinA)
= cosA + sinA = R.H.S
PlZ mark as brainliest.
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