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

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

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

prove

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

True

Solution steps

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

Manipulating left side

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

Rewrite using trig identities

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

Use the Pythagorean identity:

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

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

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

Simplify

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

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

Cancel

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

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

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 )}{csc ( x ) + 1}

Cancel the common factor:

csc (x) + 1

= csc (x) - 1

= csc (x) - 1 + 1

- 1 + 1 = 0

= csc (x)

= csc (x)

= csc (x)

We showed that the two sides could take the same form

\Rightarrow True

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

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

1 - sin (y) = \frac{cos ^{2} ( y )}{1 + sin ( y )}

\frac{cos ( θ ) cot ( θ )}{1 - sin ( θ )} - 1 = sin (θ)

1 = (sec (2 x))^{2} - (tan (2 x))^{2}

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