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Mathematics 16 Online
OpenStudy (anonymous):

prove that (2+2sinx)/cosx = 2secx ???

OpenStudy (anonymous):

what have you tried ?

OpenStudy (anonymous):

is this really \[\frac{2+2\sin(x)}{\cos(x)}=2\sec(x)\]?

OpenStudy (anonymous):

this is crazy....

OpenStudy (anonymous):

yup. so we don't need to solve it, only prove it. I had a bigger equation which I managed to simplify to the above, and I need it to equate to 2secx. I'm confused :P

OpenStudy (anonymous):

because \[\frac{2+2\sin(x)}{\cos(x)}=\frac{2}{\cos(x)}+\frac{2\sin(x)}{\cos(x)}=2\sec(x)+2\tan(x)\] and this is certainly not \[2\sec(x)\] unless of course \[\tan(x)=0\]

OpenStudy (anonymous):

maybe you could post the original problem, because this is not correct

OpenStudy (anonymous):

I had a bigger equation which I managed to simplify to the above, and I need it to equate to 2secx .... you should just type the whole equation ... hjahaha

OpenStudy (anonymous):

okayy, lemme post the original question, okayy??? \[cosx/(1-sinx) + cosx/(1+sinx) = 2secx\] Prove this.

OpenStudy (anonymous):

how'd i guess?

OpenStudy (anonymous):

not bad, broo.

OpenStudy (anonymous):

numerator is \[\cos(x)(1+\sin(x))+\cos(x)(1-\sin(x))=2\cos(x)\] denominator is \[1-\sin^2(x)=\cos^2(x)\]and then you are done by canceling

OpenStudy (anonymous):

This is too easy, you should really try it out.

OpenStudy (anonymous):

YUCK. I multiplied both the numerators with 1 + sinx, and not 2 different ones. YUCK. SLAP ME, SOMEONE.

OpenStudy (anonymous):

xD

OpenStudy (anonymous):

Slap slap slap ..

OpenStudy (anonymous):

THANK YOU.

OpenStudy (anonymous):

I got the answer. its easy. NCERT exercise type problem. To Prove: cos x/ (1 - sin x) + cos x/(1+ sin x )= 2 sec x Proof: Cross multiplying, we get, (cos x + cos x sin x + cos x - cos x sin x)/[ (1- sin x)(1 + sin x)] = 2 cos x/(1 - sin square x)= 2 cos x/ cos x cos x= 2/ cos x = 2 sec x= RHS Hence Proved.

OpenStudy (anonymous):

Thanks FFM. Cheers

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