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

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

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

prove

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

True

Solution steps

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

Manipulating left side

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

Expand

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

(- 1 + cot (x))^{2}

Apply Perfect Square Formula:

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

a = - 1, b = cot (x)

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

Simplify

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

(- 1)^{2} + 2 (- 1) cot (x) + cot^{2} (x)

Remove parentheses:

(- a) = - a

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

(- 1)^{2} = 1

(- 1)^{2}

Apply exponent rule:

(- a)^{n} = a^{n},

n

is even

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

= 1^{2}

Apply rule

1^{a} = 1

= 1

2 \cdot 1 \cdot cot (x) = 2 cot (x)

2 \cdot 1 \cdot cot (x)

Multiply the numbers:

2 \cdot 1 = 2

= 2 cot (x)

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

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

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

Manipulating right side

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

Rewrite using trig identities

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

Use the Pythagorean identity:

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

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

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

We showed that the two sides could take the same form

\Rightarrow True

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

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

\frac{1}{tan ( θ )} = cot (θ)

tan (2 x) cos (2 x) = sin (2 x)

tan (x) = 2 sin (x)

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