i need help solving this problem
(seca-tana)^2=1-sina/1+sina
any help would be appreciated
Hello, speedy229!
$\displaystyle (\sec\theta-\tan\theta)^2\:=\:\frac{1-\sin\theta}{1+\sin\theta}$
Multiply the right side by $\displaystyle \frac{1-\sin\theta}{1-\sin\theta}\!:$
. . $\displaystyle \frac{1-\sin\theta}{1+\sin\theta}\cdot{\color{blue}\frac{1-\sin\theta}{1-\sin\theta}} \;\;=\;\;\frac{(1-\sin\theta)^2}{1-\sin^2\!\theta} \;\;=\;\;\frac{(1-\sin\theta)^2}{\cos^2\!\theta}$
. . $\displaystyle = \;\;\left(\frac{1-\sin\theta}{\cos\theta}\right)^2 \;\;=\;\;\left(\frac{1}{\cos\theta} - \frac{\sin\theta}{\cos\theta}\right)^2 \;\;=\;\;(\sec\theta - \tan\theta)^2$
I think what you mean is that the answer in the back of your book has
(1-sina)/(1+sina)= (1-sina)/(1+sina) as the proof for the identities.
If that is the case start with the left side of the equation (seca-tana)^2 and use your identities to work towards the (1-sina)/(1+sina).
Soroban just proved the identity by working on the right side first, I'm sure your instructor will tell you that either method is a valid solution.