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

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

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prove sec(b)cot(b)=csc(b)

prove

sec (b) cot (b) = csc (b)

True

Solution steps

sec (b) cot (b) = csc (b)

Manipulating left side

sec (b) cot (b)

Express with sin, cos

cot (b) sec (b)

Use the basic trigonometric identity:

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

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

Use the basic trigonometric identity:

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

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

Simplify

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

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

Multiply fractions:

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

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

Cancel the common factor:

cos (b)

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

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

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

Rewrite using trig identities

Use the basic trigonometric identity:

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

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

Simplify

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

Apply the fraction rule:

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

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

Apply rule

\frac{a}{1} = a

= csc (b)

csc (b)

csc (b)

We showed that the two sides could take the same form

\Rightarrow True

\frac{1 - cos ( 2 θ )}{sin ( 2 θ )} = tan (θ)

\frac{cos ( 3 x )}{sin ( x )} + \frac{sin ( 3 x )}{cos ( x )} = 2 cot (2 x)

sin (A + B) = sin (A) + sin (B)

\frac{cot ^{2} ( B ) - cos ^{2} ( B )}{csc ^{2} ( B ) - 1} = cos^{2} (B)

cos (2 a) = 1 - 2 sin^{2} (a)