Is tan^2(θ)-sin^2(θ)=sin^4(θ)sec^2(θ) ?

The answer to whether tan^2(θ)-sin^2(θ)=sin^4(θ)sec^2(θ) is True

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

prove tan^2(θ)-sin^2(θ)=sin^4(θ)sec^2(θ)

Solution

prove

tan^{2} (θ) - sin^{2} (θ) = sin^{4} (θ) sec^{2} (θ)

Solution

True

Solution steps

tan^{2} (θ) - sin^{2} (θ) = sin^{4} (θ) sec^{2} (θ)

Manipulating left side

tan^{2} (θ) - sin^{2} (θ)

Express with sin, cos

- sin^{2} (θ) + tan^{2} (θ)

Use the basic trigonometric identity:

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

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

Simplify

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

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

Apply exponent rule:

(\frac{a}{b})^{c} = \frac{a ^{c}}{b ^{c}}

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

Convert element to fraction:

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

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

Since the denominators are equal, combine the fractions:

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

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

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

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

Factor

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

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

Factor

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

sin^{2} (θ) - cos^{2} (θ) sin^{2} (θ)

Factor out common term

sin^{2} (θ)

= sin^{2} (θ) (1 - cos^{2} (θ))

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

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

Rewrite using trig identities

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

Use the Pythagorean identity:

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

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

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

sin^{2} (θ) sin^{2} (θ) = sin^{4} (θ)

sin^{2} (θ) sin^{2} (θ)

Apply exponent rule:

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

sin^{2} (θ) sin^{2} (θ) = sin^{2 + 2} (θ)

= sin^{2 + 2} (θ)

Add the numbers:

2 + 2 = 4

= sin^{4} (θ)

= \frac{sin ^{4} ( θ )}{cos ^{2} ( θ )}

= \frac{sin ^{4} ( θ )}{cos ^{2} ( θ )}

Rewrite using trig identities

Use the basic trigonometric identity:

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

\frac{sin ^{4} ( θ )}{( \frac{1}{s e c ( θ )} ) ^{2}}

Simplify

\frac{sin ^{4} ( θ )}{( \frac{1}{s e c ( θ )} ) ^{2}}

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

(\frac{1}{sec ( θ )})^{2}

Apply exponent rule:

(\frac{a}{b})^{c} = \frac{a ^{c}}{b ^{c}}

= \frac{1 ^{2}}{sec ^{2} ( θ )}

Apply rule

1^{a} = 1

1^{2} = 1

= \frac{1}{sec ^{2} ( θ )}

= \frac{sin ^{4} ( θ )}{\frac{1}{s e c ^{2} ( θ )}}

Apply the fraction rule:

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

= \frac{sin ^{4} ( θ ) sec ^{2} ( θ )}{1}

Apply rule

\frac{a}{1} = a

= sin^{4} (θ) sec^{2} (θ)

sin^{4} (θ) sec^{2} (θ)

sin^{4} (θ) sec^{2} (θ)

We showed that the two sides could take the same form

\Rightarrow True

Popular Examples

prove cos(-θ)=cos(θ)prove

cos (- θ) = cos (θ)

prove 1+sin(2θ)=(sin(θ)+cos(θ))^2 prove

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

prove cos((5pi)/4-x)=-(sqrt(2))/2 (cos(x)+sin(x))prove

cos (\frac{5 π}{4} - x) = - \frac{2}{2} (cos (x) + sin (x))

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

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

prove 5cos^2(θ)+3sin^2(θ)=3+2cos^2(θ)prove

5 cos^{2} (θ) + 3 sin^{2} (θ) = 3 + 2 cos^{2} (θ)

Frequently Asked Questions (FAQ)

Is tan^2(θ)-sin^2(θ)=sin^4(θ)sec^2(θ) ?
The answer to whether tan^2(θ)-sin^2(θ)=sin^4(θ)sec^2(θ) is True