Is (cot(θ)+1)^2=csc^2(θ)+2cot(θ) ?

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

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prove (cot(θ)+1)^2=csc^2(θ)+2cot(θ)

prove

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

True

Solution steps

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

Manipulating left side

(cot (θ) + 1)^{2}

Expand

(1 + cot (θ))^{2} : 1 + 2 cot (θ) + cot^{2} (θ)

(1 + cot (θ))^{2}

Apply Perfect Square Formula:

(a + b)^{2} = a^{2} + 2 ab + b^{2}

a = 1, b = cot (θ)

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

Simplify

1^{2} + 2 \cdot 1 \cdot cot (θ) + cot^{2} (θ) : 1 + 2 cot (θ) + cot^{2} (θ)

1^{2} + 2 \cdot 1 \cdot cot (θ) + cot^{2} (θ)

Apply rule

1^{a} = 1

1^{2} = 1

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

Multiply the numbers:

2 \cdot 1 = 2

= 1 + 2 cot (θ) + cot^{2} (θ)

= 1 + 2 cot (θ) + cot^{2} (θ)

= 1 + cot^{2} (θ) + 2 cot (θ)

Manipulating right side

csc^{2} (θ) + 2 cot (θ)

Rewrite using trig identities

csc^{2} (θ) + 2 cot (θ)

Use the Pythagorean identity:

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

= 1 + cot^{2} (θ) + 2 cot (θ)

= 1 + cot^{2} (θ) + 2 cot (θ)

We showed that the two sides could take the same form

\Rightarrow True

cos (α - β) - cos (α + β) = 2 sin (α) sin (β)

tan (x) - 10 = \frac{sin ( x ) + 10}{cos ( x )}

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

sin (x) cos (x) = \frac{1}{tan ( x ) + cot ( x )}

1 = sec^{2} (θ) - tan^{2} (θ)

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