Is sin^2(θ)=1-1/(sec^2(θ)) ?

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

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prove sin^2(θ)=1-1/(sec^2(θ))

prove

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

True

Solution steps

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

Manipulating right side

1 - \frac{1}{sec ^{2} ( θ )}

Express with sin, cos

1 - \frac{1}{sec ^{2} ( θ )}

Use the basic trigonometric identity:

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

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

Simplify

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

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

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

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

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

(\frac{1}{cos ( θ )})^{2}

Apply exponent rule:

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

= \frac{1 ^{2}}{cos ^{2} ( θ )}

Apply rule

1^{a} = 1

1^{2} = 1

= \frac{1}{cos ^{2} ( θ )}

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

Apply the fraction rule:

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

= \frac{cos ^{2} ( θ )}{1}

Apply rule

\frac{a}{1} = a

= cos^{2} (θ)

= 1 - cos^{2} (θ)

= 1 - cos^{2} (θ)

= 1 - cos^{2} (θ)

Rewrite using trig identities

1 - cos^{2} (θ)

Use the Pythagorean identity:

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

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

= sin^{2} (θ)

= sin^{2} (θ)

We showed that the two sides could take the same form

\Rightarrow True

tan (θ) sec (θ) csc (θ) = 1 + tan^{2} (θ)

\frac{cos ( t )}{1 - sin ( t )} - tan (t) = sec (t)

\frac{1 - cos ( x )}{\frac{1}{c o s ( x )} - 1} = cos (x)

sin (π - 0) = sin (0)

cos^{2} (\frac{0}{2}) = sec (0)

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