Solve.

$\mathrm{tan}1/2x=\mathrm{csc}x-\mathrm{cot}x$

$\mathrm{tan}1/2x=\mathrm{csc}x-\mathrm{cot}x$

Raegan Bray
2022-07-29
Answered

Solve.

$\mathrm{tan}1/2x=\mathrm{csc}x-\mathrm{cot}x$

$\mathrm{tan}1/2x=\mathrm{csc}x-\mathrm{cot}x$

You can still ask an expert for help

Alden Holder

Answered 2022-07-30
Author has **15** answers

$({\mathrm{cos}}^{2}x{)}^{2}+1-({\mathrm{sin}}^{2}x{)}^{2}=2{\mathrm{cos}}^{2}x$

identity: ${\mathrm{cos}}^{2}x+{\mathrm{sin}}^{2}x=1so{\mathrm{sin}}^{2}x=1-{\mathrm{cos}}^{2}x$

$({\mathrm{cos}}^{2}x{)}^{2}+1-(1-{\mathrm{cos}}^{2}x{)}^{2}=2{\mathrm{cos}}^{2}x$

$({\mathrm{cos}}^{2}x{)}^{2}+1-[1-2{\mathrm{cos}}^{2}x+{\mathrm{cos}}^{4}x]=2{\mathrm{cos}}^{2}x$

${\mathrm{cos}}^{4}x+1-1+2{\mathrm{cos}}^{2}x-{\mathrm{cos}}^{4}x=2{\mathrm{cos}}^{2}x$

Cancel: $2{\mathrm{cos}}^{2}x=2{\mathrm{cos}}^{2}x$

identity: ${\mathrm{cos}}^{2}x+{\mathrm{sin}}^{2}x=1so{\mathrm{sin}}^{2}x=1-{\mathrm{cos}}^{2}x$

$({\mathrm{cos}}^{2}x{)}^{2}+1-(1-{\mathrm{cos}}^{2}x{)}^{2}=2{\mathrm{cos}}^{2}x$

$({\mathrm{cos}}^{2}x{)}^{2}+1-[1-2{\mathrm{cos}}^{2}x+{\mathrm{cos}}^{4}x]=2{\mathrm{cos}}^{2}x$

${\mathrm{cos}}^{4}x+1-1+2{\mathrm{cos}}^{2}x-{\mathrm{cos}}^{4}x=2{\mathrm{cos}}^{2}x$

Cancel: $2{\mathrm{cos}}^{2}x=2{\mathrm{cos}}^{2}x$

Ishaan Booker

Answered 2022-07-31
Author has **2** answers

${\mathrm{cos}}^{4}x+1-(1-{\mathrm{cos}}^{2}x)2=2{\mathrm{cos}}^{2}x,{\mathrm{cos}}^{4}x+1-(1+{\mathrm{cos}}^{4}x-2{\mathrm{cos}}^{2}x)=2{\mathrm{cos}}^{2}x$

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