Is csc(x)+cot^2(x)csc(x)=csc^3(x) ?

The answer to whether csc(x)+cot^2(x)csc(x)=csc^3(x) is True

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

prove csc(x)+cot^2(x)csc(x)=csc^3(x)

prove

csc (x) + cot^{2} (x) csc (x) = csc^{3} (x)

True

Solution steps

csc (x) + cot^{2} (x) csc (x) = csc^{3} (x)

Manipulating left side

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

Rewrite using trig identities

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

Use the Pythagorean identity:

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

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

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

Simplify

csc (x) + (csc^{2} (x) - 1) csc (x) : csc^{3} (x)

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

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

Expand

csc (x) (csc^{2} (x) - 1) : csc^{3} (x) - csc (x)

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

Apply the distributive law:

a (b - c) = ab - a c

a = csc (x), b = csc^{2} (x), c = 1

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

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

Simplify

csc^{2} (x) csc (x) - 1 \cdot csc (x) : csc^{3} (x) - csc (x)

csc^{2} (x) csc (x) - 1 \cdot csc (x)

csc^{2} (x) csc (x) = csc^{3} (x)

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

Apply exponent rule:

a^{b} \cdot a^{c} = a^{b + c}

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

= csc^{2 + 1} (x)

Add the numbers:

2 + 1 = 3

= csc^{3} (x)

1 \cdot csc (x) = csc (x)

1 \cdot csc (x)

Multiply:

1 \cdot csc (x) = csc (x)

= csc (x)

= csc^{3} (x) - csc (x)

= csc^{3} (x) - csc (x)

= csc (x) + csc^{3} (x) - csc (x)

Add similar elements:

csc (x) - csc (x) = 0

= csc^{3} (x)

= csc^{3} (x)

= csc^{3} (x)

We showed that the two sides could take the same form

\Rightarrow True

cot (x) = cos (x) arcsin (x)

\frac{1}{sin ( x ) sec ( x )} = cot (x)

3 + sin^{2} (x) = 4 - cos^{2} (x)

sin^{4} (m) - cos^{4} (m) = 1 - 2 cos^{2} (m)

\frac{\frac{1}{c o s ( x )} - 1}{1 - cos ( x )} = sec (x)

Is csc(x)+cot^2(x)csc(x)=csc^3(x) ?
The answer to whether csc(x)+cot^2(x)csc(x)=csc^3(x) is True