Is sec(x)-tan(x)*sin(x)= 1/(sec(x)) ?

The answer to whether sec(x)-tan(x)*sin(x)= 1/(sec(x)) is True

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

prove sec(x)-tan(x)*sin(x)= 1/(sec(x))

prove

sec (x) - tan (x) \cdot sin (x) = \frac{1}{sec ( x )}

True

Solution steps

sec (x) - tan (x) sin (x) = \frac{1}{sec ( x )}

Manipulating left side

sec (x) - tan (x) sin (x)

Express with sin, cos

sec (x) - sin (x) tan (x)

Use the basic trigonometric identity:

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

= \frac{1}{cos ( x )} - sin (x) tan (x)

Use the basic trigonometric identity:

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

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

Simplify

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

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

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

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

Multiply fractions:

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

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

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

sin (x) sin (x)

Apply exponent rule:

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

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

= sin^{1 + 1} (x)

Add the numbers:

1 + 1 = 2

= sin^{2} (x)

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

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

Apply rule

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

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

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

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

Rewrite using trig identities

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

Use the Pythagorean identity:

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

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

= \frac{cos ^{2} ( x )}{cos ( x )}

Cancel the common factor:

cos (x)

= cos (x)

= cos (x)

Use the basic trigonometric identity:

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

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

We showed that the two sides could take the same form

\Rightarrow True

sin^{4} (x) + cos^{4} (x) = 1 - \frac{1}{2} sin^{2} (2 x)

\frac{( 1 + cos ( 2 y ) )}{sin ^{2} ( y )} = 2 cot^{2} (y)

- tan (t) - \frac{cos ( t )}{sin ( t ) - 1} = sec (t)

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

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

Is sec(x)-tan(x)*sin(x)= 1/(sec(x)) ?
The answer to whether sec(x)-tan(x)*sin(x)= 1/(sec(x)) is True