Is sin(x)tan(x)sec(x)=sec^2(x)-1 ?

The answer to whether sin(x)tan(x)sec(x)=sec^2(x)-1 is True

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prove sin(x)tan(x)sec(x)=sec^2(x)-1

prove

sin (x) tan (x) sec (x) = sec^{2} (x) - 1

True

Solution steps

sin (x) tan (x) sec (x) = sec^{2} (x) - 1

Manipulating left side

sin (x) tan (x) sec (x)

Rewrite using trig identities

sin (x) tan (x) sec (x)

Express with sin, cos

sec (x) sin (x)

Use the basic trigonometric identity:

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

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

Simplify

\frac{1}{cos ( x )} sin (x) : \frac{sin ( x )}{cos ( x )}

\frac{1}{cos ( x )} sin (x)

Multiply fractions:

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

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

Multiply:

1 \cdot sin (x) = sin (x)

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

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

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

Use the basic trigonometric identity:

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

= tan (x) tan (x)

Simplify

tan (x) tan (x) : tan^{2} (x)

tan (x) tan (x)

Apply exponent rule:

a^{b} \cdot a^{c} = a^{b + c}

tan (x) tan (x) = tan^{1 + 1} (x)

= tan^{1 + 1} (x)

Add the numbers:

1 + 1 = 2

= tan^{2} (x)

= tan^{2} (x)

Use the Pythagorean identity:

tan^{2} (x) + 1 = sec^{2} (x)

tan^{2} (x) = sec^{2} (x) - 1

= sec^{2} (x) - 1

= sec^{2} (x) - 1

We showed that the two sides could take the same form

\Rightarrow True

tan^{4} (k) - sec^{4} (k) = 1 - 2 sec^{2} (k)

tan^{2} (x) + 5 = sec^{2} (x) + 4

1 - tan^{4} (x) = 2 sec^{2} (x) - sec^{4} (x)

sin (π) = sin (- π)

csc^{2} (x) + 2 cot (x) = (1 + cot (x))^{2}

Is sin(x)tan(x)sec(x)=sec^2(x)-1 ?
The answer to whether sin(x)tan(x)sec(x)=sec^2(x)-1 is True