Is cot(a)sec(a)=csc(a) ?

The answer to whether cot(a)sec(a)=csc(a) is True

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prove cot(a)sec(a)=csc(a)

prove

cot (a) sec (a) = csc (a)

True

Solution steps

cot (a) sec (a) = csc (a)

Manipulating left side

cot (a) sec (a)

Express with sin, cos

cot (a) sec (a)

Use the basic trigonometric identity:

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

= \frac{cos ( a )}{sin ( a )} sec (a)

Use the basic trigonometric identity:

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

= \frac{cos ( a )}{sin ( a )} \cdot \frac{1}{cos ( a )}

Simplify

\frac{cos ( a )}{sin ( a )} \cdot \frac{1}{cos ( a )} : \frac{1}{sin ( a )}

\frac{cos ( a )}{sin ( a )} \cdot \frac{1}{cos ( a )}

Multiply fractions:

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

= \frac{cos ( a ) \cdot 1}{sin ( a ) cos ( a )}

Cancel the common factor:

cos (a)

= \frac{1}{sin ( a )}

= \frac{1}{sin ( a )}

= \frac{1}{sin ( a )}

Rewrite using trig identities

Use the basic trigonometric identity:

sin (x) = \frac{1}{csc ( x )}

\frac{1}{\frac{1}{c s c ( a )}}

Simplify

\frac{1}{\frac{1}{c s c ( a )}}

Apply the fraction rule:

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

= \frac{csc ( a )}{1}

Apply rule

\frac{a}{1} = a

= csc (a)

csc (a)

csc (a)

We showed that the two sides could take the same form

\Rightarrow True

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

tan (\frac{x}{2}) + cot (\frac{x}{2}) = 2 csc (x)

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

sin (\frac{3 π}{2} + x) = - cos (x)

\frac{cos ^{2} ( t )}{sin ( t )} = csc (t) - sin (t)