Is tan(θ)sin(θ)=sec(θ)-cos(θ) ?

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

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prove tan(θ)sin(θ)=sec(θ)-cos(θ)

prove

tan (θ) sin (θ) = sec (θ) - cos (θ)

True

Solution steps

tan (θ) sin (θ) = sec (θ) - cos (θ)

Manipulating left side

tan (θ) sin (θ)

Express with sin, cos

sin (θ) tan (θ)

Use the basic trigonometric identity:

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

= sin (θ) \frac{sin ( θ )}{cos ( θ )}

Simplify

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

sin (θ) \frac{sin ( θ )}{cos ( θ )}

Multiply fractions:

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

= \frac{sin ( θ ) sin ( θ )}{cos ( 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)

= \frac{sin ^{2} ( θ )}{cos ( θ )}

= \frac{sin ^{2} ( θ )}{cos ( θ )}

= \frac{sin ^{2} ( θ )}{cos ( θ )}

Rewrite using trig identities

\frac{sin ^{2} ( θ )}{cos ( θ )}

Use the Pythagorean identity:

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

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

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

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

Manipulating right side

sec (θ) - cos (θ)

Express with sin, cos

- cos (θ) + sec (θ)

Use the basic trigonometric identity:

sec (x) = \frac{1}{cos ( x )}

= - cos (θ) + \frac{1}{cos ( θ )}

Simplify

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

- cos (θ) + \frac{1}{cos ( θ )}

Convert element to fraction:

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

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

Since the denominators are equal, combine the fractions:

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

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

- cos (θ) cos (θ) + 1 = - cos^{2} (θ) + 1

- cos (θ) cos (θ) + 1

cos (θ) cos (θ) = cos^{2} (θ)

cos (θ) cos (θ)

Apply exponent rule:

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

cos (θ) cos (θ) = cos^{1 + 1} (θ)

= cos^{1 + 1} (θ)

Add the numbers:

1 + 1 = 2

= cos^{2} (θ)

= - cos^{2} (θ) + 1

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

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

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

We showed that the two sides could take the same form

\Rightarrow True

\frac{sec ( θ )}{tan ( θ )} = csc (θ)

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

1 + tan^{2} (a) = sec^{2} (a)

2 cos (x) cos (y) = cos (x + y) + cos (x - y)

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

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