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

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

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prove ((cot^2(x)))/(1+csc(x))=csc(x)-1

prove

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

True

Solution steps

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

Manipulating left side

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

Rewrite using trig identities

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

Use the Pythagorean identity:

1 + cot^{2} (x) = csc^{2} (x)

cot^{2} (x) = csc^{2} (x) - 1

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

Simplify

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

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

Factor

csc^{2} (x) - 1 : (csc (x) + 1) (csc (x) - 1)

csc^{2} (x) - 1

Rewrite

1

1^{2}

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

Apply Difference of Two Squares Formula:

x^{2} - y^{2} = (x + y) (x - y)

csc^{2} (x) - 1^{2} = (csc (x) + 1) (csc (x) - 1)

= (csc (x) + 1) (csc (x) - 1)

= \frac{( csc ( x ) + 1 ) ( csc ( x ) - 1 )}{1 + csc ( x )}

Cancel the common factor:

csc (x) + 1

= csc (x) - 1

= csc (x) - 1

= csc (x) - 1

We showed that the two sides could take the same form

\Rightarrow True

cos (θ) = - \frac{7}{10}

sin (u) = \frac{1}{4}

tan (θ) + cot (θ) - sec (θ) csc (θ) = 0

\frac{sin ( 6 x ) - sin ( 2 x )}{cos ( 6 x ) - cos ( 2 x )} = - cot (4 x)

1 + cot^{2} (a) = csc^{2} (a)

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