Question

Can someone please help me simplify this? ((tan^2x)(csc^2x)-1))/((cscx)(tan^2x)(sinx))

Answer 1

((tan^2x)(csc^2x)-1))/((cscx)(tan^2x)= sinx

Answer 2

\[\frac{ \left( \tan ^{2}x \right)\left( \csc ^{2}x-1 \right) }{\csc x \tan ^{2}x }\]
\[\csc ^{2}x-\cot ^{2}x=1,\csc ^{2}x-1=\cot ^{2}x=\frac{ 1 }{ \tan ^{2}x }\]
\[=\frac{ \cot ^{2}x }{\csc x }=\frac{ \frac{ \cos ^{2}x }{\sin ^{2}x } }{ \frac{ 1 }{ \sin x } }\]
\[=\frac{ \cos ^{2}x }{\sin ^{2}x }*\frac{ \sin x }{ 1 }\]
solve and get the solution.

Answer 3

\[\frac{( (\frac{ \sin ^{2}x }{ \cos^{2}x } )(\frac{ 1 }{ \sin^{2}x })-1))}{ \frac{ 1 }{ sinx }(\frac{ \sin^{2}x }{ \cos^{2}x }) }\]

\[\frac{ \sec^2x-1 }{ \frac{ sinx }{ con^2x } }\]
\[\frac{ \frac{ \sin^2x }{ \cos^2x }}{ \frac{ sinx }{ \cos^2x } }\]
\[\frac{ \sin^2x }{ \cos^2x } * \frac{ \cos^2x }{ sinx } = sinx\]

Answer 4

I hope this helps :)

Answer 5

\[\frac{ \left( \tan ^{2}x \right)\left( \csc ^{2}x-1 \right) }{ \csc x \tan ^{2}x \sin x }\]
\[=\frac{ \cot ^{2}x }{\frac{ 1 }{\sin x }\sin x }=\cot ^{2}x\]

Answer 6

I have verified my answer before posting it. after I simplified everything using trig identities I arrived at "sin x"

Answer 7

i failed to understand how you posted sec^2 x-1