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Someone help me with proving identities? and please show the work if you can!

Someone help me with proving identities? and please show the work if you can!-example-1

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Answer:

See Below.

Explanation:

11)

We have:


\displaystyle \cos\theta\cdot \cot\theta+\sin^2\theta\cdot \csc\theta=\csc\theta

Rewrite:


\displaystyle \cos\theta\left((\cos\theta)/(\sin\theta)\right)+\sin^2\theta\left((1)/(\sin\theta)}\right)=\csc\theta

Multiply:


\displaystyle (\cos^2\theta)/(\sin\theta)+(\sin^2\theta)/(\sin\theta)=\csc\theta

Combine:


\displaystyle (\cos^2\theta+\sin^2\theta)/(\sin\theta)=\csc\theta

According to the Pythagorean Identity, sin²θ + cos²θ = 1. Hence:


\displaystyle (1)/(\sin\theta)=\csc\theta

Simplify:


\csc\theta=\csc\theta

12)

We have:


\displaystyle (\cos^2\theta +\sin^2\theta)/(1+\tan^2\theta)=\cos^2\theta

Pythagorean Identity:


\displaystyle (1)/(1+\tan^2\theta)=\cos^2\theta

From the Pythagorean Identity, if we divide both sides by cos²θ, we acquire that tan²θ + 1 = sec²θ. Hence:


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

Simplify:


\cos^2\theta =\cos^2\theta

13)

We have:


\displaystyle (\csc\theta\cdot \cos\theta)/(\tan\theta +\cot\theta)=\cos^2\theta

Rewrite:


\displaystyle (1/\sin\theta\cdot \cos\theta)/(\sin\theta/\cos\theta+\cos\theta/\sin\theta)=\cos^2\theta

Multiply both top and bottom by cosθsinθ. Hence:


\displaystyle (\cos^2\theta)/(\sin^2\theta +\cos^2\theta)=\cos^2\theta

By the Pythagorean Identity:


\displaystyle (\cos^2\theta)/(1)=\cos^2\theta

Simplify:


\cos^2\theta=\cos^2\theta

User Jim Garvin
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