Question

Proof trigonometry identities.\csc(x)+\cot(x)=\frac{1}{\csc(x)-\cot(x)}

Trigonometric equation and identitie
ANSWERED
asked 2021-09-08
Proof trigonometry identities.
\(\displaystyle{\csc{{\left({x}\right)}}}+{\cot{{\left({x}\right)}}}={\frac{{{1}}}{{{\csc{{\left({x}\right)}}}-{\cot{{\left({x}\right)}}}}}}\)

Expert Answers (1)

2021-09-09
Let as consider the given function,
L.H.S.=R.H.S.
First to solve R.H.S.,
\(\displaystyle{\frac{{{1}}}{{{\csc{{\left({x}\right)}}}-{\cot{{\left({x}\right)}}}}}}\)
Use the following trigonometric identity: \(\displaystyle{1}=-{{\cot}^{{2}}{\left({x}\right)}}+{{\csc}^{{2}}{\left({x}\right)}}\)
\(\displaystyle{\frac{{{1}}}{{{\csc{{\left({x}\right)}}}-{\cot{{\left({x}\right)}}}}}}={\frac{{-{{\cot}^{{2}}{\left({x}\right)}}+{{\csc}^{{2}}{\left({x}\right)}}}}{{-{\cot{{\left({x}\right)}}}+{\csc{{\left({x}\right)}}}}}}\)
Apply Difference of Two Squares Formula:
\(\displaystyle{{\csc}^{{2}}{\left({x}\right)}}-{{\cot}^{{2}}{\left({x}\right)}}={\left({\csc{{\left({x}\right)}}}+{\cot{{\left({x}\right)}}}\right)}{\left({\csc{{\left({x}\right)}}}-{\cot{{\left({x}\right)}}}\right)}\)
\(\displaystyle{\frac{{{1}}}{{{\csc{{\left({x}\right)}}}-{\cot{{\left({x}\right)}}}}}}={\frac{{{\left({\csc{{\left({x}\right)}}}+{\cot{{\left({x}\right)}}}\right)}{\left({\csc{{\left({x}\right)}}}-{\cot{{\left({x}\right)}}}\right)}}}{{-{\cot{{\left({x}\right)}}}+{\csc{{\left({x}\right)}}}}}}\)
\(\displaystyle{\frac{{{1}}}{{{\csc{{\left({x}\right)}}}-{\cot{{\left({x}\right)}}}}}}={\csc{{\left({x}\right)}}}+{\cot{{\left({x}\right)}}}\)
Hence, LHS=RHS
10
 
Best answer

expert advice

Have a similar question?
We can deal with it in 3 hours
...