Is cot(b)sec(b)sin(b)=1 ?

The answer to whether cot(b)sec(b)sin(b)=1 is True

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prove cot(b)sec(b)sin(b)=1

prove

cot (b) sec (b) sin (b) = 1

True

Solution steps

cot (b) sec (b) sin (b) = 1

Manipulating left side

cot (b) sec (b) sin (b)

Express with sin, cos

cot (b) sec (b) sin (b)

Use the basic trigonometric identity:

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

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

Use the basic trigonometric identity:

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

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

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

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

Multiply fractions:

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

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

Cancel the common factor:

cos (b)

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

Cancel the common factor:

sin (b)

= 1

= 1

= 1

We showed that the two sides could take the same form

\Rightarrow True

\frac{( sec ^{2} ( Q ) - 1 )}{sec ^{2} ( Q )} = sin^{2} (Q)

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

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

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

\frac{1 - 4 sec ^{2} ( x )}{1 + 2 sec ( x )} = 1 - 2 sec (x)