This is part of practice, and I am clueless.
I have to prove: $\dfrac{1}{\cos θ} - \cos θ = \sin θ \tan θ$
Using the identities: $\sin^2 θ + \cos^2 θ = 1$, and/or $\tan θ = \dfrac{\sin θ}{\cos θ}$.
This is part of practice, and I am clueless.
I have to prove: $\dfrac{1}{\cos θ} - \cos θ = \sin θ \tan θ$
Using the identities: $\sin^2 θ + \cos^2 θ = 1$, and/or $\tan θ = \dfrac{\sin θ}{\cos θ}$.
Using the identity:
$\tan{\theta}\equiv\frac{\sin{\theta}}{\cos{\theta}}$
And multiplying both sides by $\sin{\theta}$:
$\sin{\theta}\tan{\theta}\equiv\frac{\sin^{2}{\theta}}{\cos{\theta}}\equiv\frac{1-\cos^{2}{\theta}}{\cos{\theta}}\equiv\frac{1}{\cos{\theta}}-\cos{\theta} \qquad\blacksquare$
Hint: $\frac{1}{\cos\theta}-\cos\theta=\frac{1-\cos^2\theta}{\cos\theta}= \dots.$