Prove that sinA/1-cosA =1 +cosA/sinA  

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jeyaram's profile pic

jeyaram | Student, Undergraduate | (Level 1) Valedictorian

Posted on

sinA/1-cosA =1 +cosA/sinA

L.H.S= sinA/1-cosA

=sinA(1+cosA)/(1-cosA)(1+cosA)

=sinA(1+cosA)/1-cos^2(A)

=sinA(1+cosA)/sin^2(A)

=1 +cosA/sinA

=R.H.S

krishna-agrawala's profile pic

krishna-agrawala | College Teacher | (Level 3) Valedictorian

Posted on

We have to prove:

sinA/(1 - cosA) = (1 + cosA)/sinA

To do this we multiply both sides of equation by (1 - cosA)*sinA, giving

[sinA/(1 - cosA)]*[(1 - cosA)*sinA] = [(1 + cosA)/sinA]*[(1 - cosA)*sinA]

Therefor:

sinA^2* = (1 + cosA)*(1 - cosA)

Therefore:

sinA^2* = 1 - cosA^2

Shifting the term cosA^2 from right hand side to left hand side in the above equation we get:

sinA^2* + cosA^2 = 1

We know above relationship to be true. Therefor the given equation is true.

neela's profile pic

neela | High School Teacher | (Level 3) Valedictorian

Posted on

To prove sinA/(1-cosA) = (1+cosA)/sinA.

Proof:

we know that,

sin^2 A+cos^2 A= 1, is a trigonometric identity.

Therefore,

sin^2 A = 1- cos^2 A  = (1+cosA)(1-cosA). So,

Sin^2 A /(1-cosA) = 1+cosA  or

sinA/(1-cosA) = (1+cosA)/sinA.

 

Note:

The original problem is  sinA/1-cosA = 1+cosA/sinA. This is not an identity, for,

SinA/1-cosA  = 1+cosA/sinA. This means:

Put A= 0 deg. Then,

LHS = sin 0 /1 -cos0) = 1/1 - 1 = 1 -1 = 0 .

RHS= 1 + cos 0 / sin 0 = 1+1/0 = infinite.

Therefore, it is edited to read as:

Prove sinA/(1-cosA) = (1+cosA)/sinA.

 

 

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