Trigonometric identities transformation last one

[Drain Brain]

Transform the left hand member into the right hand member.

$\frac{\tan\alpha+\tan\beta}{\sec\alpha-\sec\beta}=\frac{\sec\alpha+\sec\beta}{\tan\alpha-\tan\beta}$By using cross multiplication I was able to prove this identity but what I actually want to accomplush is to transform the left member to the right member and that I have no clue on. I need some help here.

Thanks!

 

[Dethrone]

Hi (Wave),

On the LHS, multiply by the conjugate $\sec\alpha+\sec\beta$ and on the RHS, multiply by the conjugate $\tan\alpha+\tan\beta$. Without expanding anything, see if you can simplify their denominators so that they are the same.

 

[Drain Brain]

Hi (Wave),

On the LHS, multiply by the conjugate $\sec\alpha+\sec\beta$ and on the RHS, multiply by the conjugate $\tan\alpha+\tan\beta$. Without expanding anything, see if you can simplify their denominators so that they are the same.

Did you mean multiply the conjugate by the numerator and denominator of both members?

 

[Dethrone]

Yep. Multiply the LHS numerator and denominator by $\sec\alpha+\sec\beta$, and similarly, in the RHS, multiply numerator and denominator by $\tan\alpha+\tan\beta$. Don't expand the numerator after that, and factor the denominator via difference of squares.

 

[MarkFL]

When I was a student, proving trig. identities meant taking the left side, and through the use of certain basic identities, transforming the left side into the right.

\(\displaystyle \frac{\tan\alpha+\tan\beta}{\sec\alpha-\sec\beta}\)

I think you are off to a good start by trying:

\(\displaystyle \frac{\tan\alpha+\tan\beta}{\sec\alpha-\sec\beta}\cdot\frac{\tan\alpha-\tan\beta}{\tan\alpha-\tan\beta}\cdot\frac{\sec\alpha+\sec\beta}{\sec\alpha+\sec\beta}\)

You should then write it in the form:

\(\displaystyle \frac{\tan^2\alpha-\tan^2\beta}{\sec^2\alpha-\sec^2\beta}\cdot\frac{\sec\alpha+\sec\beta}{\tan\alpha-\tan\beta}\)

Now, on the first factor, apply a Pythagorean identity to either the numerator or denominator...

 

[Prove It]

When I was a student, proving trig. identities meant taking the left side, and through the use of certain basic identities, transforming the left side into the right.

Or rather, taking EITHER side, and through the use of certain basic identities, transforming that side into the OTHER side...

 

[Fantini]

You can use that

$$\tan^2(\alpha) - \sec^2 (\alpha) = -1,$$

$$\tan^2(\beta) - \sec^2(\beta) = -1,$$

therefore

$$\tan^2(\alpha) - \sec^2(\alpha) = \tan^2(\beta) - \sec^2(\beta).$$

Can you take it from here?

:)