Answered

Explore Westonci.ca, the leading Q&A site where experts provide accurate and helpful answers to all your questions. Explore our Q&A platform to find reliable answers from a wide range of experts in different fields. Get precise and detailed answers to your questions from a knowledgeable community of experts on our Q&A platform.

How do you do,
csc2x-cot2x = tanx

Sagot :

iloveonedirection
[tex]csc2x-cot2x=tanx \\ \\ (\frac{1}{2}cscxsecx) -( \frac{cotx-tanx}{2} )=tanx \\ \\ (\frac{cscxsecx}{2}) -( \frac{cotx-tanx}{2} )=tanx \\ \\ (\frac{ \frac{1}{sinx} \frac{1}{cosx} }{2}) -( \frac{cotx-tanx}{2} )=tanx \\ \\ (\frac{1}{2(sinxcosx)}) -( \frac{cotx-tanx}{2} )=tanx \\ \\ (\frac{1}{2(sinxcosx)}) -( \frac{(sinxcosx)*cotx-tanx}{2*(sinxcosx)} )=tanx \\ \\ \frac{1-(cos^2x-sin^2x)}{2(sinxcosx)} =tanx \\ \\ \frac{1-cos^2+sin^2x}{2(sinxcosx)}=tanx [/tex]

[tex]\frac{sin^2x+sin^2x}{2(sinxcosx)} =tanx \\ \\ \frac{1-cos2x}{sin2x} =tanx \\ \\ \frac{sin2x}{cos2x+1}=tanx \\ \\ tanx=tanx [/tex]


Thanks for using our platform. We're always here to provide accurate and up-to-date answers to all your queries. We appreciate your visit. Our platform is always here to offer accurate and reliable answers. Return anytime. Westonci.ca is your trusted source for answers. Visit us again to find more information on diverse topics.