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

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

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

prove

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

True

Solution steps

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

Manipulating left side

sec (x) (sec (x) - cos (x))

Expand

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

(- cos (x) + sec (x)) sec (x)

= sec (x) (- cos (x) + sec (x))

Apply the distributive law:

a (b + c) = ab + a c

a = sec (x), b = - cos (x), c = sec (x)

= sec (x) (- cos (x)) + sec (x) sec (x)

Apply minus-plus rules

+ (- a) = - a

= - sec (x) cos (x) + sec (x) sec (x)

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

sec (x) sec (x)

Apply exponent rule:

a^{b} \cdot a^{c} = a^{b + c}

sec (x) sec (x) = sec^{1 + 1} (x)

= sec^{1 + 1} (x)

Add the numbers:

1 + 1 = 2

= sec^{2} (x)

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

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

Rewrite using trig identities

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

sec (x) cos (x) = 1

sec (x) cos (x)

Express with sin, cos

sec (x) cos (x)

Use the basic trigonometric identity:

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

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

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

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

Multiply fractions:

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

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

Cancel the common factor:

cos (x)

= 1

= 1

= sec^{2} (x) - 1

Use the Pythagorean identity:

sec^{2} (x) = tan^{2} (x) + 1

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

= tan^{2} (x)

= tan^{2} (x)

We showed that the two sides could take the same form

\Rightarrow True

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

sin (3 x) = 3 sin (x) - 4 (sin (x))^{3}

(5 sin (θ) - 5 cos (θ))^{2} = 25 - 25 sin (2 θ)

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

cot (x) \cdot sec (x) = csc (x)

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