# Please solve:1+sinA/cosA = cosA/1-sinA

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1+sinA/cosA = cosA/1-sinA

L.H.S

=1+sinA/cosA

=cosA(secA+tanA)/{(1-sinA)[(cosA(1+sinA)/(1-sinA)(1+sinA)]}

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

=cosA/1-sinA

=R.H.S

### Please solve:

1+sinA/cosA = cosA/1-sinA

There is a difference between 1+sinA/cosA = cosA/1-sinA and (1+sinA)/cosA =cosA/(1-sinA).

I take the second case first and show that it is an **identity**.Hence it holds **for all values of x .**

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

Take the right side:cosA/(1-sinA).

Multiply numerator and denominator by 1+sinA which does not alter the expression value and simplify and show whether it is equal to the left side expression. If it is equal, then it ia an identity.

cosA(1+sinA)/{1-sinA)(1+sinA)}

=cosA(1+sinA)/{1-(sinA)^2}, as (cosA)^2+(sinA)^2=1

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

=(1+sinA)/cosA = left left side.

Threfore , **(1+sinA)/cosA = cosA/(1-sinA)** is an identity and so, holds good for all values of A.

Part (ii):

Now let us take your actual problem:

Solve: ** 1+sinA/cosA=cosA/1-sinA **. Now we solve this equation . This is not an identity. So , the equation can be solved for A. Since this is not an identity, the solution may be for a particular value/s of for which only the equation holds good.

The given equation is exactly equivalent to (by the PEDMA order of priority) : 1+(sinA/cosA) = (cosA/1) - sinA and you cannot mean or interpret otherwise.

Substitute tanA=x, then sinA =x/sqrt(1+x^2) , and cosA=1/sqrt(1+x^2).The given equation , therefore, becomes:

1+x=1/sqrt(1+x^2) + x/sqrt(1+x^2)

Multiply both sides by sqrt(1+x^2) :

sqrt(1+x^2)(1+x)=1-x

square both sides and simplify :

(1+x^2)(1+x)^2=(1-x)^2

(1+x^2)(1+2x+x^2)=1-2x+x^2

1+2x+2x^2+2x^3+x^4=1-2x+x^2.

4x+x^2+2x^3+x^4=0

x(4+x+2x^2+x^3)=0

Therefore x=0, or

4+x+2x^2+x^3=0, which is cubic equation with one real value , x=-2.314596213 , by iteration in exel**.**Therefore, **tanA = 0** or **tan A= -2.314596213**** **

Threfore A = tan inverse (0) or

A =Tan inverse (-2.314596213)

**A=0 degre**e. or A=-66.63368553 deg or **A = 113.3663145** degree.

Therefore, the solution for A : A=0 or A =113.3663145. The other solution is an extraneous solution.

Tally :

When A=**0**, equation (1) gives:

1+tan0=cos0-sin0: LHS=1+0=1 and RHS=1-0=1.

When A=**113.3663145**

LHS:1+tan(**113.3663145**)= **-1.314596212.....**

RHS:cos(**113.3663145**)-sin(**113.3663145**)=**-1.314596212..**

Hope this helps.

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

Therefore:[(1+sinA)/cosA]*[cosA*(1-sinA)]

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

Therefore: (1+sinA)*(1-sinA) = cosA*cosA

Therefore: [1-(sinA)^2] = (cosA)^2

Therefore: 1 = [(cosA)^2]+[(sinA)^2]

we know that [(cosA)^2]+[(sinA)^2] = 1

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