Is tan^2(θ)csc^2(θ)=sec^2(θ) ?

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

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prove tan^2(θ)csc^2(θ)=sec^2(θ)

prove

tan^{2} (θ) csc^{2} (θ) = sec^{2} (θ)

True

Solution steps

tan^{2} (θ) csc^{2} (θ) = sec^{2} (θ)

Manipulating left side

tan^{2} (θ) csc^{2} (θ)

Express with sin, cos

csc^{2} (θ) tan^{2} (θ)

Use the basic trigonometric identity:

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

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

Use the basic trigonometric identity:

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

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

Simplify

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

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

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

(\frac{1}{sin ( θ )})^{2}

Apply exponent rule:

(\frac{a}{b})^{c} = \frac{a ^{c}}{b ^{c}}

= \frac{1 ^{2}}{sin ^{2} ( θ )}

Apply rule

1^{a} = 1

1^{2} = 1

= \frac{1}{sin ^{2} ( θ )}

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

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

(\frac{sin ( θ )}{cos ( θ )})^{2}

Apply exponent rule:

(\frac{a}{b})^{c} = \frac{a ^{c}}{b ^{c}}

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

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

Multiply fractions:

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

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

Cancel the common factor:

sin^{2} (θ)

= \frac{1}{cos ^{2} ( θ )}

= \frac{1}{cos ^{2} ( θ )}

= \frac{1}{cos ^{2} ( θ )}

Rewrite using trig identities

Use the basic trigonometric identity:

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

\frac{1}{( \frac{1}{s e c ( θ )} ) ^{2}}

Simplify

\frac{1}{( \frac{1}{s e c ( θ )} ) ^{2}}

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

(\frac{1}{sec ( θ )})^{2}

Apply exponent rule:

(\frac{a}{b})^{c} = \frac{a ^{c}}{b ^{c}}

= \frac{1 ^{2}}{sec ^{2} ( θ )}

Apply rule

1^{a} = 1

1^{2} = 1

= \frac{1}{sec ^{2} ( θ )}

= \frac{1}{\frac{1}{s e c ^{2} ( θ )}}

Apply the fraction rule:

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

= \frac{sec ^{2} ( θ )}{1}

Apply rule

\frac{a}{1} = a

= sec^{2} (θ)

sec^{2} (θ)

sec^{2} (θ)

We showed that the two sides could take the same form

\Rightarrow True

(1 + tan (x)) (1 - cot (x)) = tan (x) - cot (x)

(tan (x) + cot (x))^{2} = sec^{4} (x) cot^{2} (x)

(1 + tan^{2} (s)) (1 - sin^{2} (s)) = 1

8 - 4 csc^{2} (z) = 4 - 4 cot^{2} (z)

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

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