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

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

User AndreyKo
by
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4 votes

Answer:

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.

User Asfand Shabbir
by
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