Question

prove the following,

cosA÷1-tanA+sinA÷1-cotA=cotA+sinA​

Answer 1

Answer:

here is your answer

Step-by-step explanation:

How do I prove the following identity, (cosA)/(1-tanA)+(SinA)/(1-cotA)=sinA+cosA?

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This is easy.

First, manipulate the LHS

cos(A)1−tan(A)+sin(A)1−cot(A)

Express in terms of sin and cos

=cos(A)1−sin(A)cos(A)+sin(A)1−cos(A)sin(A)

Let us now simplify cos(A)1−sin(A)cos(A)

For the denominator 1−sin(A)cos(A) , we convert this to =cos(A)−sin(A)cos(A)

resulting to =cos(A)cos(A)−sin(A)cos(A)

Applying the fraction rule abc=a⋅cb

We have =cos(A)cos(A)cos(A)−sin(A)→=cos2(A)cos(A)−sin(A)

For sin(A)1−cos(A)sin(A) , we use the same processes as above, resulting in

=sin(A)sin(A)−cos(A)sin(A)→=sin(A)sin(A)sin(A)−cos(A)→=sin2(A)sin(A)−cos(A)

So now we go back to this equation: =cos2(A)cos(A)−sin(A)+sin2(A)sin(A)−cos(A)

Using the LCD cos(A)−sin(A) , we have this equation:

=cos2(A)cos(A)−sin(A)−sin2(A)cos(A)−sin(A)→cos2(A)−sin2(A)cos(A)−sin(A)

Using the difference of squares rule: x2−y2=(x+y)(x−y)

So now we have =(sin(A)+cos(A))(cos(A)−sin(A))cos(A)−sin(A)

Cancel the common factor cos(A)−sin(A)

We are now left with =sin(A)+cos(A)

∴cos(A)1−tan(A)+sin(A)1−cot(A)=sin(A)+cos(A): TRUE