Prove that:- cosecA/cotA+tanA=cosA?

Answer 1

So you want to prove that

cscA/[cotA + tanA] = cosA

Since the left hand side (or LHS) is more complex, we're going to choose to work wtih that.

LHS = cscA/[cotA + tanA]

Changing everything to sines and cosines,

LHS = [1/sinA] / [ (cosA/sinA) + (sinA/cosA) ]

What we want to do is eliminate all fractions within the fraction by multiplying top and bottom by sinAcosA.

LHS = cosA / [ cos^2(A) + sin^2(A) ]

Notice that the bottom is the identity sin^2(x) + cos^2(x) = 1. We change it into that.

LHS = cosA / 1

And anything divided by 1 is itself, so

LHS = cosA = RHS

Answer 2

Answer:

From L.H.S
Numerator = cosec A = 1 / sin A

Denominator = cotA+tanA = (cos A / sin A) + (sin A / cos A)
Taking L.C.M and combining the terms,
= (cos^2 A + sin^2 A) / sin A cos A
= 1 / sin A cos A
Numerator / Denominator = (1 / sin A) * sin A cos A
= cos A = R.H.S

Answer 3

cosec A / cot A + tan A = 1 / cos A + sin A / cos A

= 1+sin A / cos A

Use brackets dammit,
then
cosec A / (cot A + tan A) =cos A.sin A / [sin A (sin^2 A + cos^2 A)]
= cos A
As required.

Answer 4

cosecA/(cotA+tanA)=cosA
LHS
= cosecA/(cosA/sinA + sinA/cosA)
= cosecA/(1/sinAcosA)
= 1/sinA X cosA/sinA
= cosA
= RHS

Answer 5

i imagine you may first amplify the (cosecA - cotA)^2, to get something like a^2 - 2ab + b^2, then back, cosec and cot must be amplify to cos and sin. attempt to amplify it first and it is going to lead you to the ans.

Answer 6

it must be cosecA/(cotA+tanA)=cosA
taking L.H.S
cosecA/(cotA+tanA)
=(1/sinA) /[(cosA/sinA)+(sinA/cosA)]
taking the lcm in denominator
=(1/sinA)/[(cos^2A+sin^2A)/(sinAcosA)]
=(1/sinA)/(1/sinAcosA)
=(1/sinA)sinAcosA
=cosA=R.h.s

Answer 7

1/sinA/cosA/sinA + sinA/cosA = cosA
1/sinA/sin2A+cos2A/sinAcosA=cosA
1/sinA/1/sinAcosA=cosA
cosA=cosA