Is sin^2(θ)sec(θ)csc(θ)=tan(θ) ?

The answer to whether sin^2(θ)sec(θ)csc(θ)=tan(θ) is True

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prove sin^2(θ)sec(θ)csc(θ)=tan(θ)

prove

sin^{2} (θ) sec (θ) csc (θ) = tan (θ)

True

Solution steps

sin^{2} (θ) sec (θ) csc (θ) = tan (θ)

Manipulating left side

sin^{2} (θ) sec (θ) csc (θ)

Express with sin, cos

csc (θ) sec (θ) sin^{2} (θ)

Use the basic trigonometric identity:

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

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

Use the basic trigonometric identity:

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

= \frac{1}{sin ( θ )} \cdot \frac{1}{cos ( θ )} sin^{2} (θ)

Simplify

\frac{1}{sin ( θ )} \cdot \frac{1}{cos ( θ )} sin^{2} (θ) : \frac{sin ( θ )}{cos ( θ )}

\frac{1}{sin ( θ )} \cdot \frac{1}{cos ( θ )} sin^{2} (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 sin ^{2} ( θ )}{sin ( θ ) cos ( θ )}

Multiply:

1 \cdot 1 \cdot sin^{2} (θ) = sin^{2} (θ)

= \frac{sin ^{2} ( θ )}{sin ( θ ) cos ( θ )}

Cancel the common factor:

sin (θ)

= \frac{sin ( θ )}{cos ( θ )}

= \frac{sin ( θ )}{cos ( θ )}

= \frac{sin ( θ )}{cos ( θ )}

Use the basic trigonometric identity:

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

= tan (θ)

We showed that the two sides could take the same form

\Rightarrow True

6 cot^{2} (y) (sec^{2} (y) - 1) = 6

csc (x) (cos (x) + sin (x)) = cot (x) + 1

cos^{2} (x) + cos^{2} (x) cot^{2} (x) = cot^{2} (x)

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

2 cot^{2} (A) sin^{2} (A) = 1 + cos (2 A)

Is sin^2(θ)sec(θ)csc(θ)=tan(θ) ?
The answer to whether sin^2(θ)sec(θ)csc(θ)=tan(θ) is True