Is prove-sin^2(x)sec^2(x)=1-sec^2(x) ?

The answer to whether prove-sin^2(x)sec^2(x)=1-sec^2(x) 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(x)sec^2(x)=1-sec^2(x)

Solution

prove

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

Solution

True

Solution steps

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

Manipulating left side

- sin^{2} (x) sec^{2} (x)

Rewrite using trig identities

- sin^{2} (x) sec^{2} (x)

Use the Pythagorean identity:

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

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

Use the Pythagorean identity:

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

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

= - (1 - cos^{2} (x)) (tan^{2} (x) + 1)

= - (1 - cos^{2} (x)) (tan^{2} (x) + 1)

Expand

- (1 + tan^{2} (x)) (1 - cos^{2} (x)) : - 1 + cos^{2} (x) - tan^{2} (x) + tan^{2} (x) cos^{2} (x)

- (1 + tan^{2} (x)) (1 - cos^{2} (x))

Expand

(1 + tan^{2} (x)) (1 - cos^{2} (x)) : 1 - cos^{2} (x) + tan^{2} (x) - tan^{2} (x) cos^{2} (x)

(1 + tan^{2} (x)) (1 - cos^{2} (x))

Apply FOIL method:

(a + b) (c + d) = a c + a d + b c + b d

a = 1, b = tan^{2} (x), c = 1, d = - cos^{2} (x)

= 1 \cdot 1 + 1 \cdot (- cos^{2} (x)) + tan^{2} (x) \cdot 1 + tan^{2} (x) (- cos^{2} (x))

Apply minus-plus rules

+ (- a) = - a

= 1 \cdot 1 - 1 \cdot cos^{2} (x) + 1 \cdot tan^{2} (x) - tan^{2} (x) cos^{2} (x)

Simplify

1 \cdot 1 - 1 \cdot cos^{2} (x) + 1 \cdot tan^{2} (x) - tan^{2} (x) cos^{2} (x) : 1 - cos^{2} (x) + tan^{2} (x) - tan^{2} (x) cos^{2} (x)

1 \cdot 1 - 1 \cdot cos^{2} (x) + 1 \cdot tan^{2} (x) - tan^{2} (x) cos^{2} (x)

Multiply the numbers:

1 \cdot 1 = 1

= 1 - 1 \cdot cos^{2} (x) + 1 \cdot tan^{2} (x) - tan^{2} (x) cos^{2} (x)

Multiply:

1 \cdot cos^{2} (x) = cos^{2} (x)

= 1 - cos^{2} (x) + 1 \cdot tan^{2} (x) - tan^{2} (x) cos^{2} (x)

Multiply:

1 \cdot tan^{2} (x) = tan^{2} (x)

= 1 - cos^{2} (x) + tan^{2} (x) - tan^{2} (x) cos^{2} (x)

= 1 - cos^{2} (x) + tan^{2} (x) - tan^{2} (x) cos^{2} (x)

= - (1 - cos^{2} (x) + tan^{2} (x) - tan^{2} (x) cos^{2} (x))

Distribute parentheses

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

Apply minus-plus rules

- (- a) = a, - (a) = - a

= - 1 + cos^{2} (x) - tan^{2} (x) + tan^{2} (x) cos^{2} (x)

= - 1 + cos^{2} (x) - tan^{2} (x) + cos^{2} (x) tan^{2} (x)

Rewrite using trig identities

- 1 + cos^{2} (x) - tan^{2} (x) + cos^{2} (x) tan^{2} (x)

Use the Pythagorean identity:

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

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

= cos^{2} (x) + cos^{2} (x) tan^{2} (x) - sec^{2} (x)

Use the basic trigonometric identity:

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

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

Simplify

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

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

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

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

Apply exponent rule:

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

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

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

Multiply fractions:

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

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

Cancel the common factor:

cos^{2} (x)

= sin^{2} (x)

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

Use the Pythagorean identity:

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

= 1 - sec^{2} (x)

= 1 - sec^{2} (x)

We showed that the two sides could take the same form

\Rightarrow True

Popular Examples

prove 1/(1+cos^2(x))+1/(1+sec^2(x))=1 prove

\frac{1}{1 + cos ^{2} ( x )} + \frac{1}{1 + sec ^{2} ( x )} = 1

prove sin^2(θ)+sin^2(θ)tan^2(θ)=tan^2(θ)prove

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

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

1 + sin (2 θ) = (cos (θ) sin (θ))^{2}

prove sec(θ)csc(θ)=tan(θ)+cot(θ)prove

sec (θ) csc (θ) = tan (θ) + cot (θ)

prove 2tan(0)=sin(0)prove

2 tan (0^{\circ}) = sin (0^{\circ})

Frequently Asked Questions (FAQ)

Is prove-sin^2(x)sec^2(x)=1-sec^2(x) ?
The answer to whether prove-sin^2(x)sec^2(x)=1-sec^2(x) is True