Is sec(x)cos(x)-2cos^2(x)=-(cos(2x)) ?

The answer to whether sec(x)cos(x)-2cos^2(x)=-(cos(2x)) is True

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prove sec(x)cos(x)-2cos^2(x)=-(cos(2x))

prove

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

True

Solution steps

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

Manipulating left side

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

Express with sin, cos

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

Use the basic trigonometric identity:

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

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

Simplify

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

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

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

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

Multiply fractions:

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

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

Cancel the common factor:

cos (x)

= 1

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

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

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

Rewrite using trig identities

1 - 2 cos^{2} (x)

Use the Double Angle identity:

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

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

= - cos (2 x)

= - cos (2 x)

We showed that the two sides could take the same form

\Rightarrow True

2 cos^{2} (A) - cos (2 A) - 1 = 0

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

\frac{cot ( x ) + 1}{cos ( x ) + sin ( x )} = csc (x)

sec^{2} (y) = (sec (y))^{2}

csc (A) = \frac{7}{4}

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