Is tan(θ)(csc(θ)-sin(θ))=cos(θ) ?

The answer to whether tan(θ)(csc(θ)-sin(θ))=cos(θ) is True

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prove tan(θ)(csc(θ)-sin(θ))=cos(θ)

prove

tan (θ) (csc (θ) - sin (θ)) = cos (θ)

True

Solution steps

tan (θ) (csc (θ) - sin (θ)) = cos (θ)

Manipulating left side

tan (θ) (csc (θ) - sin (θ))

Express with sin, cos

(csc (θ) - sin (θ)) tan (θ)

Use the basic trigonometric identity:

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

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

Use the basic trigonometric identity:

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

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

Simplify

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

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

Multiply fractions:

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

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

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 )}{cos ( θ )}

Multiply

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

sin (θ) \frac{1 - sin ^{2} ( 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)

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

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

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

Rewrite using trig identities

\frac{1 - sin ^{2} ( θ )}{cos ( θ )}

Use the Pythagorean identity:

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

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

= \frac{cos ^{2} ( θ )}{cos ( θ )}

Cancel the common factor:

cos (θ)

= cos (θ)

= cos (θ)

We showed that the two sides could take the same form

\Rightarrow True

cot^{2} (x) sec^{2} (x) = csc^{2} (x)

sin (10 θ) = 2 sin (5 θ) cos (5 θ)

\frac{cos ( θ ) sec ( θ )}{tan ( θ )} = cot (θ)

\frac{cos ( θ ) - 1}{sin ( θ )} = cot (θ) - csc (θ)

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

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