Is sec(A)-cos(A)=tan(A)sin(A) ?

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

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prove sec(A)-cos(A)=tan(A)sin(A)

prove

sec (A) - cos (A) = tan (A) sin (A)

True

Solution steps

sec (A) - cos (A) = tan (A) sin (A)

Manipulating left side

sec (A) - cos (A)

Express with sin, cos

- cos (A) + sec (A)

Use the basic trigonometric identity:

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

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

Simplify

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

- cos (A) + \frac{1}{cos ( A )}

Convert element to fraction:

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

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

Since the denominators are equal, combine the fractions:

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

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

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

- cos (A) cos (A) + 1

cos (A) cos (A) = cos^{2} (A)

cos (A) cos (A)

Apply exponent rule:

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

cos (A) cos (A) = cos^{1 + 1} (A)

= cos^{1 + 1} (A)

Add the numbers:

1 + 1 = 2

= cos^{2} (A)

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

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

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

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

Manipulating right side

tan (A) sin (A)

Express with sin, cos

sin (A) tan (A)

Use the basic trigonometric identity:

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

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

Simplify

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

sin (A) \frac{sin ( A )}{cos ( A )}

Multiply fractions:

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

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

sin (A) sin (A) = sin^{2} (A)

sin (A) sin (A)

Apply exponent rule:

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

sin (A) sin (A) = sin^{1 + 1} (A)

= sin^{1 + 1} (A)

Add the numbers:

1 + 1 = 2

= sin^{2} (A)

= \frac{sin ^{2} ( A )}{cos ( A )}

= \frac{sin ^{2} ( A )}{cos ( A )}

= \frac{sin ^{2} ( A )}{cos ( A )}

Rewrite using trig identities

\frac{sin ^{2} ( A )}{cos ( A )}

Use the Pythagorean identity:

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

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

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

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

We showed that the two sides could take the same form

\Rightarrow True

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

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

cos (θ) tan (- θ) = - sin (θ)

tan^{3} (x) cot (x) = sec^{2} (x) - 1

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

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