Is (sin^2(t))/(cos(t))=sec(t)-cos(t) ?

The answer to whether (sin^2(t))/(cos(t))=sec(t)-cos(t) is True

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

prove (sin^2(t))/(cos(t))=sec(t)-cos(t)

prove

\frac{sin ^{2} ( t )}{cos ( t )} = sec (t) - cos (t)

True

Solution steps

\frac{sin ^{2} ( t )}{cos ( t )} = sec (t) - cos (t)

Manipulating right side

sec (t) - cos (t)

Express with sin, cos

- cos (t) + sec (t)

Use the basic trigonometric identity:

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

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

Simplify

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

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

Convert element to fraction:

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

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

Since the denominators are equal, combine the fractions:

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

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

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

- cos (t) cos (t) + 1

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

cos (t) cos (t)

Apply exponent rule:

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

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

= cos^{1 + 1} (t)

Add the numbers:

1 + 1 = 2

= cos^{2} (t)

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

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

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

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

Rewrite using trig identities

\frac{1 - cos ^{2} ( t )}{cos ( t )}

Use the Pythagorean identity:

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

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

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

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

We showed that the two sides could take the same form

\Rightarrow True

tan (θ) + cot (θ) = 2 csc (2 θ)

csc^{4} (x) - cot^{4} (x) = 2 cot^{2} (x) + 1

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

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

cos (- x) - sin (- x) = cos (x) + sin (x)

Is (sin^2(t))/(cos(t))=sec(t)-cos(t) ?
The answer to whether (sin^2(t))/(cos(t))=sec(t)-cos(t) is True