Is 1-(tan(θ)cos(θ))/(csc(θ))=cos^2(θ) ?

The answer to whether 1-(tan(θ)cos(θ))/(csc(θ))=cos^2(θ) is True

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prove 1-(tan(θ)cos(θ))/(csc(θ))=cos^2(θ)

prove

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

True

Solution steps

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

Manipulating left side

1 - \frac{tan ( θ ) cos ( θ )}{csc ( θ )}

Express with sin, cos

1 - \frac{cos ( θ ) tan ( θ )}{csc ( θ )}

Use the basic trigonometric identity:

tan (x) = \frac{sin ( x )}{cos ( x )}

= 1 - \frac{cos ( θ ) \frac{s i n ( θ )}{c o s ( θ )}}{csc ( θ )}

Use the basic trigonometric identity:

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

= 1 - \frac{cos ( θ ) \frac{s i n ( θ )}{c o s ( θ )}}{\frac{1}{s i n ( θ )}}

Simplify

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

1 - \frac{cos ( θ ) \frac{s i n ( θ )}{c o s ( θ )}}{\frac{1}{s i n ( θ )}}

\frac{cos ( θ ) \frac{s i n ( θ )}{c o s ( θ )}}{\frac{1}{s i n ( θ )}} = sin^{2} (θ)

\frac{cos ( θ ) \frac{s i n ( θ )}{c o s ( θ )}}{\frac{1}{s i n ( θ )}}

Apply the fraction rule:

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

= \frac{cos ( θ ) \frac{s i n ( θ )}{c o s ( θ )} sin ( θ )}{1}

Apply the fraction rule:

\frac{a}{1} = a

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

Multiply fractions:

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

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

Cancel the common factor:

cos (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)

= 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

Is 1-(tan(θ)cos(θ))/(csc(θ))=cos^2(θ) ?
The answer to whether 1-(tan(θ)cos(θ))/(csc(θ))=cos^2(θ) is True