Is (csc(θ)-sin(θ))/(csc(θ))=cos^2(θ) ?

The answer to whether (csc(θ)-sin(θ))/(csc(θ))=cos^2(θ) is True

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

prove (csc(θ)-sin(θ))/(csc(θ))=cos^2(θ)

prove

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

True

Solution steps

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

Manipulating left side

\frac{csc ( θ ) - sin ( θ )}{csc ( θ )}

Express with sin, cos

\frac{csc ( θ ) - sin ( θ )}{csc ( θ )}

Use the basic trigonometric identity:

csc (x) = \frac{1}{sin ( x )}

= \frac{\frac{1}{s i n ( θ )} - sin ( θ )}{\frac{1}{s i n ( θ )}}

Simplify

\frac{\frac{1}{s i n ( θ )} - sin ( θ )}{\frac{1}{s i n ( θ )}} : 1 - sin^{2} (θ)

\frac{\frac{1}{s i n ( θ )} - sin ( θ )}{\frac{1}{s i n ( θ )}}

Apply the fraction rule:

\frac{a}{\frac{b}{c}} = \frac{a \cdot c}{b}

= \frac{( \frac{1}{s i n ( θ )} - sin ( θ ) ) sin ( θ )}{1}

Join

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

\frac{1}{sin ( θ )} - sin (θ)

Convert element to fraction:

sin (θ) = \frac{sin ( θ ) sin ( θ )}{sin ( θ )}

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

Since the denominators are equal, combine the fractions:

\frac{a}{c} \pm \frac{b}{c} = \frac{a \pm b}{c}

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

1 - sin (θ) sin (θ) = 1 - sin^{2} (θ)

1 - sin (θ) sin (θ)

sin (θ) sin (θ) = sin^{2} (θ)

sin (θ) sin (θ)

Apply exponent rule:

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

sin (θ) sin (θ) = sin^{1 + 1} (θ)

= sin^{1 + 1} (θ)

Add the numbers:

1 + 1 = 2

= sin^{2} (θ)

= 1 - sin^{2} (θ)

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

= \frac{\frac{- s i n ^{2} ( θ ) + 1}{s i n ( θ )} sin ( θ )}{1}

Apply the fraction rule:

\frac{a}{1} = a

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

Multiply fractions:

a \cdot \frac{b}{c} = \frac{a \cdot b}{c}

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

Cancel the common factor:

sin (θ)

= 1 - sin^{2} (θ)

= 1 - sin^{2} (θ)

= 1 - sin^{2} (θ)

Rewrite using trig identities

1 - sin^{2} (θ)

Use the Pythagorean identity:

1 = cos^{2} (x) + sin^{2} (x)

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

= cos^{2} (θ)

= cos^{2} (θ)

We showed that the two sides could take the same form

\Rightarrow True

\frac{sin ( θ ) sec ( θ )}{tan ( θ )} = 1

\frac{tan ^{2} ( θ )}{tan ^{2} ( θ ) + 1} = sin^{2} (θ)

tan (\frac{π}{6}) = \frac{1}{3}

cos (4 a) = 1 - 8 sin^{2} (a) cos^{2} (a)

(cos (x)) (tan (x) + cot (x)) = csc (x)

Is (csc(θ)-sin(θ))/(csc(θ))=cos^2(θ) ?
The answer to whether (csc(θ)-sin(θ))/(csc(θ))=cos^2(θ) is True