Is (1+sin(z))(1-sin(z))= 1/(sec^2(z)) ?

The answer to whether (1+sin(z))(1-sin(z))= 1/(sec^2(z)) is True

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

prove (1+sin(z))(1-sin(z))= 1/(sec^2(z))

prove

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

True

Solution steps

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

Manipulating right side

\frac{1}{sec ^{2} ( z )}

Express with sin, cos

\frac{1}{sec ^{2} ( z )}

Use the basic trigonometric identity:

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

= \frac{1}{( \frac{1}{c o s ( z )} ) ^{2}}

Simplify

\frac{1}{( \frac{1}{c o s ( z )} ) ^{2}} : cos^{2} (z)

\frac{1}{( \frac{1}{c o s ( z )} ) ^{2}}

(\frac{1}{cos ( z )})^{2} = \frac{1}{cos ^{2} ( z )}

(\frac{1}{cos ( z )})^{2}

Apply exponent rule:

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

= \frac{1 ^{2}}{cos ^{2} ( z )}

Apply rule

1^{a} = 1

1^{2} = 1

= \frac{1}{cos ^{2} ( z )}

= \frac{1}{\frac{1}{c o s ^{2} ( z )}}

Apply the fraction rule:

\frac{1}{\frac{b}{c}} = \frac{c}{b}

= \frac{cos ^{2} ( z )}{1}

Apply rule

\frac{a}{1} = a

= cos^{2} (z)

= cos^{2} (z)

= cos^{2} (z)

Rewrite using trig identities

cos^{2} (z)

Use the Pythagorean identity:

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

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

= 1 - sin^{2} (z)

Factor

1 - sin^{2} (z) : (1 + sin (z)) (1 - sin (z))

1 - sin^{2} (z)

Apply Difference of Two Squares Formula:

z^{2} - y^{2} = (z + y) (z - y)

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

= (1 + sin (z)) (1 - sin (z))

= (1 + sin (z)) (1 - sin (z))

= (1 + sin (z)) (1 - sin (z))

We showed that the two sides could take the same form

\Rightarrow True

\frac{tan ( x )}{sec ( x ) + 1} = csc (x) - cot (x)

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

\frac{( 1 + tan ( θ ) )}{1 + cot ( θ )} = tan (θ)

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

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

Is (1+sin(z))(1-sin(z))= 1/(sec^2(z)) ?
The answer to whether (1+sin(z))(1-sin(z))= 1/(sec^2(z)) is True