Is 1/(sec^2(x))=sin^2(x)+cos(2x) ?

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

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

prove

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

True

Solution steps

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

Manipulating left side

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

Express with sin, cos

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

Use the basic trigonometric identity:

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

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

Simplify

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

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

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

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

Apply exponent rule:

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

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

Apply rule

1^{a} = 1

1^{2} = 1

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

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

Apply the fraction rule:

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

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

Apply rule

\frac{a}{1} = a

= cos^{2} (x)

= cos^{2} (x)

= cos^{2} (x)

Rewrite using trig identities

cos^{2} (x)

Use the Double Angle identity:

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

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

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

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

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

We showed that the two sides could take the same form

\Rightarrow True

csc (2 y) = csc (y) 7

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

tan (x) + cot (2 x) = sec (2 x)

csc (θ) (csc (θ) - sin (θ)) = cot^{2} (θ)

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

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