Need help

[native brook]

Solve for theta

[neon blade]

For these problems turning everything into sin and cos usually helps.

[native brook]

How do I do it without that

It’s not allowed

For this homework

[neon blade]

What are you allowed to do? I don’t see any other way to algebraically solve this

[copper willow]

My next guess would be to square both sides. And then use Pythag identity

[maiden flare]

$\theta={\frac{\pi}{2}+2\pi k, \frac{4\pi}{3}+2\pi k}$

[solemn edge]

Lorenzo Borchardt

[opaque vapor]

pi/2 can't be a solution as both tan and sec are then undefined

Are you allowed to use the identity $\sec\theta-\tan\theta\sin\theta=\cos\theta$?

[solemn edge]

Jay

[opaque vapor]

It can be provd from the standard Pythagoream identities...
\begin{align*} \sin^2{\theta} + \cos^2{\theta} &= 1 \ \tan^2{\theta} + 1 &= \sec^2{\theta} \ \sec^2{\theta} - \tan^2{\theta} &= 1 \ \sec^2{\theta}\cos\theta - \tan^2{\theta}\cos\theta &= \cos\theta \ \sec^2{\theta}\left(\sec\theta\right)^{-1} - \tan{\theta}\cdot\frac{\sin\theta}{\cos\theta}\cdot\cos\theta &= \cos\theta \ \sec{\theta} - \tan\theta\sin\theta &= \cos\theta \end{align*}

[solemn edge]

Jay

[final dust]

Just turn it into cos and sin at this point 🙏😭

Everything cancels out perfectly

[opaque vapor]

The OP said they couldn't just turn everything into sine and cosine. My way does let the problem be simplified greatly, but I'm not sure if it is allowed given the restrfictions to which the OP alluded.