Question

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

Please get this done and match the sides. Thank you!

Answer 1

See Below.

Explanation:

We want to prove the trigonometric identity:

`\displaystyle (\sec^2(\theta)-1)/(\sin(\theta))=(\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 (\sin(\theta))/(1-\sin^2(\theta))=(\sin(\theta))/(\cos^2(\theta))`

Split:

`\displaystyle =(\sin(\theta))/(\cos(\theta))\left((1)/(\cos(\theta))\right)=\tan(\theta)\sec(\theta)`

Therefore, our equation becomes:

`\displaystyle (\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 (\tan^2(\theta))/(\sin(\theta))=\tan(\theta)\sec(\theta)`

Rewrite:

`\displaystyle (\tan(\theta))^2\left((1)/(\sin(\theta))\right)=\tan(\theta)\sec(\theta)`

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

`\displaystyle (\sin^2(\theta))/(\cos^2(\theta))\left((1)/(\sin(\theta))\right)=\tan(\theta)\sec(\theta)`

Simplify:

`\displaystyle (\sin(\theta))/(\cos^2(\theta))=\tan(\theta)\sec(\theta)`

Simplify:

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

Hence proven.