Is (csc^2(x))/(csc^2(x)-1)=sec^2(x) ?

The answer to whether (csc^2(x))/(csc^2(x)-1)=sec^2(x) is True

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prove (csc^2(x))/(csc^2(x)-1)=sec^2(x)

prove

\frac{csc ^{2} ( x )}{csc ^{2} ( x ) - 1} = sec^{2} (x)

True

Solution steps

\frac{csc ^{2} ( x )}{csc ^{2} ( x ) - 1} = sec^{2} (x)

Manipulating right side

sec^{2} (x)

Express with sin, cos

sec^{2} (x)

Use the basic trigonometric identity:

sec (x) = \frac{1}{cos ( x )}

= (\frac{1}{cos ( x )})^{2}

Simplify

(\frac{1}{cos ( x )})^{2} : \frac{1}{cos ^{2} ( x )}

(\frac{1}{cos ( x )})^{2}

Apply exponent rule:

(\frac{a}{b})^{c} = \frac{a ^{c}}{b ^{c}}

= \frac{1 ^{2}}{cos ^{2} ( x )}

Apply rule

1^{a} = 1

1^{2} = 1

= \frac{1}{cos ^{2} ( x )}

= \frac{1}{cos ^{2} ( x )}

= \frac{1}{cos ^{2} ( x )}

Rewrite using trig identities

\frac{1}{cos ^{2} ( x )}

Use the Pythagorean identity:

cos^{2} (x) + sin^{2} (x) = 1

cos^{2} (x) = 1 - sin^{2} (x)

= \frac{1}{1 - sin ^{2} ( x )}

= \frac{1}{1 - sin ^{2} ( x )}

Rewrite using trig identities

Use the basic trigonometric identity:

sin (x) = \frac{1}{csc ( x )}

\frac{1}{1 - ( \frac{1}{c s c ( x )} ) ^{2}}

Simplify

\frac{1}{1 - ( \frac{1}{c s c ( x )} ) ^{2}}

(\frac{1}{csc ( x )})^{2} = \frac{1}{csc ^{2} ( x )}

(\frac{1}{csc ( x )})^{2}

Apply exponent rule:

(\frac{a}{b})^{c} = \frac{a ^{c}}{b ^{c}}

= \frac{1 ^{2}}{csc ^{2} ( x )}

Apply rule

1^{a} = 1

1^{2} = 1

= \frac{1}{csc ^{2} ( x )}

= \frac{1}{1 - \frac{1}{c s c ^{2} ( x )}}

Join

1 - \frac{1}{csc ^{2} ( x )} : \frac{csc ^{2} ( x ) - 1}{csc ^{2} ( x )}

1 - \frac{1}{csc ^{2} ( x )}

Convert element to fraction:

1 = \frac{1 csc ^{2} ( x )}{csc ^{2} ( x )}

= \frac{1 \cdot csc ^{2} ( x )}{csc ^{2} ( x )} - \frac{1}{csc ^{2} ( x )}

Since the denominators are equal, combine the fractions:

\frac{a}{c} \pm \frac{b}{c} = \frac{a \pm b}{c}

= \frac{1 \cdot csc ^{2} ( x ) - 1}{csc ^{2} ( x )}

Multiply:

1 \cdot csc^{2} (x) = csc^{2} (x)

= \frac{csc ^{2} ( x ) - 1}{csc ^{2} ( x )}

= \frac{1}{\frac{c s c ^{2} ( x ) - 1}{c s c ^{2} ( x )}}

Apply the fraction rule:

\frac{1}{\frac{b}{c}} = \frac{c}{b}

= \frac{csc ^{2} ( x )}{csc ^{2} ( x ) - 1}

\frac{csc ^{2} ( x )}{csc ^{2} ( x ) - 1}

\frac{csc ^{2} ( x )}{csc ^{2} ( x ) - 1}

We showed that the two sides could take the same form

\Rightarrow True

sin (2 x) + sin (2 y) = 2 sin (x + y) cos (x - y)

sin (θ) csc (θ) cos (θ) = cos (θ)

cot (x) - csc^{2} (x) cot (x) = - cot^{3} (x)

cot (x - \frac{π}{2}) = - tan (x)

\frac{sin ( 5 x ) + sin ( 11 x )}{cos ( 5 x ) + cos ( 11 x )} = tan (8 x)

Is (csc^2(x))/(csc^2(x)-1)=sec^2(x) ?
The answer to whether (csc^2(x))/(csc^2(x)-1)=sec^2(x) is True