Is sin(θ)cos(θ)sec(θ)csc(θ)=1 ?

The answer to whether sin(θ)cos(θ)sec(θ)csc(θ)=1 is True

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prove sin(θ)cos(θ)sec(θ)csc(θ)=1

prove

sin (θ) cos (θ) sec (θ) csc (θ) = 1

True

Solution steps

sin (θ) cos (θ) sec (θ) csc (θ) = 1

Manipulating left side

sin (θ) cos (θ) sec (θ) csc (θ)

Express with sin, cos

cos (θ) csc (θ) sec (θ) sin (θ)

Use the basic trigonometric identity:

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

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

Use the basic trigonometric identity:

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

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

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

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

Multiply fractions:

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

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

Cancel the common factor:

cos (θ)

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

Cancel the common factor:

sin (θ)

= 1 \cdot 1

Multiply the numbers:

1 \cdot 1 = 1

= 1

= 1

= 1

We showed that the two sides could take the same form

\Rightarrow True

sin (\frac{π}{2} - 0) = cos (0)

sin (3 x) = sin (3 - 4 sin^{2} (x))

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

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

cot (x) - 1 = \frac{csc ( x ) - sec ( x )}{sec ( x )}

Is sin(θ)cos(θ)sec(θ)csc(θ)=1 ?
The answer to whether sin(θ)cos(θ)sec(θ)csc(θ)=1 is True