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

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

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

prove

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

True

Solution steps

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

Manipulating left side

sin (x) tan (x)

Express with sin, cos

sin (x) tan (x)

Use the basic trigonometric identity:

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

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

Simplify

sin (x) \frac{sin ( x )}{cos ( x )} : \frac{sin ^{2} ( x )}{cos ( x )}

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

Multiply fractions:

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

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

sin (x) sin (x) = sin^{2} (x)

sin (x) sin (x)

Apply exponent rule:

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

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

= sin^{1 + 1} (x)

Add the numbers:

1 + 1 = 2

= sin^{2} (x)

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

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

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

Rewrite using trig identities

\frac{sin ^{2} ( x )}{cos ( x )}

Use the Pythagorean identity:

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

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

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

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

Use the basic trigonometric identity:

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

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

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

We showed that the two sides could take the same form

\Rightarrow True

4 \cdot sin (\frac{x}{4}) \cdot cos (\frac{x}{4}) cos (\frac{x}{2}) = sin (x)

\frac{sin ( x )}{cos ( x ) + sin ( x )} = 11

\frac{1 + cos ( a )}{sin ( a )} = cot (\frac{a}{s})

cot (x) (1 - cos (2 x)) = 2 cos^{2} (x) tan (x)

8 \cdot sin^{4} (θ) = cos (4 θ) + 4 \cdot cos (2 θ) - 3

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