[zaibjaffer]

tan3a-tan2a-tana = tan3a tan2a tana

[ebaines]

It's a bit of work, but...

Write out each tangent as the sine/cosine equivalent, and then use trig identities to convert the terms that are multiples of 'a' into sina and cosa equivalents. For example: sin2a = 2sinacosa and cos2a = cos^2a-sin^2a. I will leave it to you to figure out how to converyt sin3a and cos3a into terms of sina and cosa. Once you get it into a form that has only sina and cosa terms, you'll see it works out.

[zaibjaffer]

It's a bit of work, but...

Write out each tangent as the sine/cosine equivalent, and then use trig identities to convert the terms that are multiples of 'a' into sina and cosa equivalents. For example: sin2a = 2sinacosa and cos2a = cos^2a-sin^2a. I will leave it to you to figure out how to converyt sin3a and cos3a into terms of sina and cosa. Once you get it into a form that has only sina and cosa terms, you'll see it works out.

I an still confused... I tried doing it the way you suggested but don't get the answer... do you think this way actually works.

[ebaines]

I an still confused .....I tried doing it the way you suggested but dont get the answer......do you think this way actually works.

Yes, it does.

I started by combining the left hand side over a common denominator, which turns out to have the same denominator as the right hand side. So then it's merely a job of showing that both numerators are the same. In other words:

sin(3a)*cos(2a)*cos(a) - sin(2a)*cos(3a)*cos(a) - sin(a)*cos(3a)*cos(2a) = sin(3a)*sin(2a)*sin(a)

At this point, you substitute for sin(3a), cos(3a), sin(2a) and cos(2a) with identities that are purely in sin(a) and cos(a). Do you know what those identities are?

[zaibjaffer]

Yes, it does.

I started by combining the left hand side over a common denominator, which turns out to have the same denominator as the right hand side. So then it's merely a job of showing that both numerators are the same. In other words:

sin(3a)*cos(2a)*cos(a) - sin(2a)*cos(3a)*cos(a) - sin(a)*cos(3a)*cos(2a) = sin(3a)*sin(2a)*sin(a)

At this point, you substitute for sin(3a), cos(3a), sin(2a) and cos(2a) with identities that are purely in sin(a) and cos(a). Do you know what those identities are?

Thanks for helping me it worked...
I have an other question now
2sin5xcos4x-sinx=sin9x

[ebaines]

You shouldn't post a new question in the same thread as an old. Please start a new question.

[hohoho123456789]

Ha ha. Very funny:-)