Prove secA-cosA=tanAsinA Must show all work. HELP PLEASE?

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Math

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ishpiro's profile pic

Posted on

When working with trig identities, rewriting secant as a reciprocal of a cosine is a good place to start:

`secA-cosA=1/cosA-cosA`

`` Then, perform the subtraction by rewriting cosA as a fraction with the denominator cosA:

`1/cosA - cosA = 1/cosA - (cos^2A)/cosA=(1-cos^2A)/cosA`

`` Now use the Pythagorean identity to rewrite the numerator:

`1-cos^2A=sin^2A`

Finally, `(sin^2A)/cosA=sinA/cosA*sinA=tanAsinA`

`` which is equivalent to the right hand side of the identity you had to prove.

 

 

 

violy's profile pic

Posted on

Let us try to prove that left side is equal to the right side.

We know that secA = 1/cosA. 

So, we will have:

`1/(cosA) - cosA`  

Hence, lcd = cosA. We multiply top and bottom of cosA by cosA. 

`1/(cosA) - (cosA*cosA)/cosA = 1/(cosA) - (cos^2A)/cosA = (1 - (cos^2A))/cosA`

We know that sin^2A = 1 - cos^2A. 

`(1 - (cos^2A))/(cosA) = (sin^2A)/cosA`

We can rewrite that:

`(sin^2A)/(cosA) = (sinA/cosA)*sinA`

Take note that tanA = sinA/cosA. 

`(sinA/cosA)*sinA = tanAsinA`

 

user1450001's profile pic

Posted on

secA = 1/cosA
i.e..,
secA-cosA=[1/cosA]-cosA
=[1-cosA*cosA]/cosA
=[1-cos^2A]/cosA (since 1-cos^2A= SinA)
=sin^2A/cosA
=sinA*sinA/cosA
=[sinA/cosA]*sinA
=TanA*sinA

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