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I need help with this practice problem solving I could not get the entire problem in one photo so I will send an additional photo of the rest

I need help with this practice problem solving I could not get the entire problem-example-1
User Im
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4 votes

Answer:


(\cos x)/(1-\sin x)=(1)/(\cos x)+(\sin x)/(\cos x)
(\cos x)/(1-\sin x)=(1+\sin x)/(\cos x)
(\cos x)/(1-\sin x)=((1+\sin x)\cdot\cos x)/(1-\sin ^2x)
(\cos x)/(1-\sin x)=\frac{(1+\sin x)\cdot\cos x}{(1-\sin^{}x)(1+\sin x)}
(\cos x)/(1-\sin x)=(\cos x)/(1-\sin x)

Step-by-step explanation:

First, we realise that


\sec x=(1)/(\cos x)

and


\tan x=(\sin x)/(\cos x)

therefore,


\sec x+\tan x=(1)/(\cos x)+(\sin x)/(\cos x)

Now,


(1)/(\cos x)+(\sin x)/(\cos x)=(1+\sin x)/(\cos x)

Using the fact that


(1)/(\cos x)=(\cos x)/(\cos^2x)=(\cos x)/(1-\sin^2x)

Therefore,


(1+\sin x)/(\cos x)=(\cos x(1+\sin x))/(1-\sin^2x)
=(\cos x(1+\sin x))/((1-\sin x)(1+\sin x))=(\cos x)/(1-\sin x)

Hence, we have shown that


\boxed{\sec x+\tan x=(\cos x)/(1-\sin x)\text{.}}

User Kluge
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