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

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

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

prove

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

True

Solution steps

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

Manipulating right side

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

Express with sin, cos

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

Use the basic trigonometric identity:

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

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

Simplify

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

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

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

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

Apply exponent rule:

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

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

Apply rule

1^{a} = 1

1^{2} = 1

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

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

Apply the fraction rule:

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

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

Apply rule

\frac{a}{1} = a

= cos^{2} (a)

= cos^{2} (a)

= cos^{2} (a)

Rewrite using trig identities

cos^{2} (a)

Use the Pythagorean identity:

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

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

= 1 - sin^{2} (a)

Factor

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

1 - sin^{2} (a)

Apply Difference of Two Squares Formula:

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

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

= (1 + sin (a)) (1 - sin (a))

= (1 + sin (a)) (1 - sin (a))

= (1 + sin (a)) (1 - sin (a))

We showed that the two sides could take the same form

\Rightarrow True

sec (a) (1 - sin^{2} (a)) = cos (a)

cos (x + y) - cos (x - y) = - 2 sin (x) sin (y)

sec (\frac{π}{2} - u) = csc (u)

\frac{cot ^{3} ( t )}{csc ( t )} = cos (t) (csc^{2} (- 1))

\frac{sec ^{2} ( x ) cot ( x )}{csc ^{2} ( x )} = tan (x)

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