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

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jeyaram's profile pic

jeyaram | Student, Undergraduate | (Level 1) Valedictorian

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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

 

neela's profile pic

neela | High School Teacher | (Level 3) Valedictorian

Posted on

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 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.....

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

Hope this helps.



 

 

 

 

krishna-agrawala's profile pic

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

Posted on

(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)

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