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Question

kekeman · Answer

Its either I and III or I,II, and III

kekeman · Answer

So either option C or D

kekeman · Answer

I leaning towards C more though

surjithayer · Answer

$$\frac{ \cos 2x }{ \cos x}=\frac{ \cos ^2x-\sin ^2x }{ \cos x }=\frac{ \cos ^2x }{ \cos x }-\frac{ \sin ^2x }{ \cos x }$$
$$=\cos x-\left( \sin x \right)\left( \tan x \right)$$
so 1.
$$1+\cos 2x=2\cos ^2x$$
$$\cos 2x=2\cos ^2x-1$$
divide by cos x
$$\frac{ \cos 2x }{ \cos x }=\frac{ 2\cos ^2x }{ \cos x }-\frac{ 1 }{ \cos x}$$
$$=2\cos x-\sec x$$
so 111
answer 1 and 111
C

kekeman · Answer

Thanks you have great explanations and i love that u show your work lifesaver fr

kekeman · Answer

@surjithayer wrote:

$$\frac{ \cos 2x }{ \cos x}=\frac{ \cos ^2x-\sin ^2x }{ \cos x }=\frac{ \cos ^2x }{ \cos x }-\frac{ \sin ^2x }{ \cos x }$$ $$=\cos x-\left( \sin x \right)\left( \tan x \right)$$ so 1. $$1+\cos 2x=2\cos ^2x$$ $$\cos 2x=2\cos ^2x-1$$ divide by cos x $$\frac{ \cos 2x }{ \cos x }=\frac{ 2\cos ^2x }{ \cos x }-\frac{ 1 }{ \cos x}$$ $$=2\cos x-\sec x$$ so 111 answer 1 and 111 C

could you help me with another one