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

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

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

prove

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

True

Solution steps

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

Manipulating left side

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

Rewrite using trig identities

cot^{2} (x) + \frac{1}{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

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

Apply rule

\frac{a}{1} = a

\frac{1}{1} = 1

= cot^{2} (x) + 1

Use the Pythagorean identity:

1 + cot^{2} (x) = csc^{2} (x)

= csc^{2} (x)

= csc^{2} (x)

We showed that the two sides could take the same form

\Rightarrow True

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

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

tan^{2} (θ) = sec^{2} (θ) - 1

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

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

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