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

jeyaram | Student

1+sinA/cosA = cosA/1-sinA








neela | Student

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 .


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)^2},   as (cosA)^2+(sinA)^2=1


=(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) :


square both sides and simplify :






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 degree. 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.....


Hope this helps.





krishna-agrawala | Student

(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)

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