Is 2tan(x)sec(x)csc(x)=2+2tan^2(x) ?

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

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Popular Trigonometry >

prove 2tan(x)sec(x)csc(x)=2+2tan^2(x)

Solution

prove

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

Solution

True

Solution steps

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

Manipulating left side

2 tan (x) sec (x) csc (x)

Express with sin, cos

2 csc (x) sec (x) tan (x)

Use the basic trigonometric identity:

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

= 2 \cdot \frac{1}{sin ( x )} sec (x) tan (x)

Use the basic trigonometric identity:

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

= 2 \cdot \frac{1}{sin ( x )} \cdot \frac{1}{cos ( x )} tan (x)

Use the basic trigonometric identity:

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

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

Simplify

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

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

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 ( x ) \cdot 2}{sin ( x ) cos ( x ) cos ( x )}

Cancel the common factor:

sin (x)

= \frac{1 \cdot 1 \cdot 2}{cos ( x ) cos ( x )}

Multiply the numbers:

1 \cdot 1 \cdot 2 = 2

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

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

cos (x) cos (x)

Apply exponent rule:

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

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

= cos^{1 + 1} (x)

Add the numbers:

1 + 1 = 2

= cos^{2} (x)

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

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

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

Manipulating right side

2 + 2 tan^{2} (x)

Factor

2 + 2 tan^{2} (x) : 2 (1 + tan^{2} (x))

2 + 2 tan^{2} (x)

Rewrite as

= 2 \cdot 1 + 2 tan^{2} (x)

Factor out common term

2

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

= (1 + tan^{2} (x)) \cdot 2

Express with sin, cos

(1 + tan^{2} (x)) \cdot 2

Use the basic trigonometric identity:

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

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

Simplify

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

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

Apply exponent rule:

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

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

Join

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

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

Convert element to fraction:

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

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

Since the denominators are equal, combine the fractions:

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

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

Multiply:

1 \cdot cos^{2} (x) = cos^{2} (x)

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

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

Multiply fractions:

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

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

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

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

Rewrite using trig identities

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

Use the Pythagorean identity:

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

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

Simplify

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

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

We showed that the two sides could take the same form

\Rightarrow True

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Frequently Asked Questions (FAQ)

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