Question

What is the identity of (sec^2theta-1)/sintheta = sintheta/(1-sin^2theta)

Answer 1

Answer:

See Below.

Step-by-step explanation:

We want to prove the trigonometric identity:

$\displaystyle \frac{\sec^2(\theta)-1}{\sin(\theta)}=\frac{\sin(\theta)}{1-\sin^2(\theta)}$

To start, let's simplify the right side. Recall the Pythagorean Identity:

$\sin^2(\theta)+\cos^2(\theta)=1$

Therefore:

$\cos^2(\theta)=1-\sin^2(\theta)$

Substitute:

$\displaystyle \frac{\sin(\theta)}{1-\sin^2(\theta)}=\frac{\sin(\theta)}{\cos^2(\theta)}$

Split:

$\displaystyle =\frac{\sin(\theta)}{\cos(\theta)}\left(\frac{1}{\cos(\theta)}\right)=\tan(\theta)\sec(\theta)$

Therefore, our equation becomes:

$\displaystyle \frac{\sec^2(\theta)-1}{\sin(\theta)}=\tan(\theta)\sec(\theta)$

From the Pythagorean Identity, we can divide both sides by cos²(θ). This yields:

$\displaystyle \tan^2(\theta)+1=\sec^2(\theta)$

So:

$\tan^2(\theta)=\sec^2(\theta)-1$

Substitute:

$\displaystyle \frac{\tan^2(\theta)}{\sin(\theta)}=\tan(\theta)\sec(\theta)$

Rewrite:

$\displaystyle (\tan(\theta))^2\left(\frac{1}{\sin(\theta)}\right)=\tan(\theta)\sec(\theta)$

Recall that tan(θ) = sin(θ)/cos(θ). So:

$\displaystyle \frac{\sin^2(\theta)}{\cos^2(\theta)}\left(\frac{1}{\sin(\theta)}\right)=\tan(\theta)\sec(\theta)$

Simplify:

$\displaystyle \frac{\sin(\theta)}{\cos^2(\theta)}=\tan(\theta)\sec(\theta)$

Simplify:

$\tan(\theta)\sec(\theta)=\tan(\theta)\sec(\theta)}$

Hence proven.