Is 1+(cot^2(θ))/(csc(θ)+1)=csc(θ) ?

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

prove 1+(cot^2(θ))/(csc(θ)+1)=csc(θ)

prove

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

True

Solution steps

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

Manipulating left side

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

Rewrite using trig identities

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

Use the Pythagorean identity:

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

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

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

Simplify

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

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

Cancel

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

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

Factor

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

csc^{2} (θ) - 1

Rewrite

1

1^{2}

= csc^{2} (θ) - 1^{2}

Apply Difference of Two Squares Formula:

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

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

= (csc (θ) + 1) (csc (θ) - 1)

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

Cancel the common factor:

csc (θ) + 1

= csc (θ) - 1

= 1 + csc (θ) - 1

Group like terms

= csc (θ) + 1 - 1

1 - 1 = 0

= csc (θ)

= csc (θ)

= csc (θ)

We showed that the two sides could take the same form

\Rightarrow True

csc^{2} (x) - cos (x) sec (x) = cot^{2} (x)

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

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

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

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

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