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Prove the trigonometric identity $\sec\left(x\right)^2+\csc\left(x\right)^2=\frac{1}{\sin\left(x\right)^2\cos\left(x\right)^2}$

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Solution

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Final answer to the problem

true

Step-by-step Solution

How should I solve this problem?

  • Express everything into Sine and Cosine
  • Prove from LHS (left-hand side)
  • Prove from RHS (right-hand side)
  • Exact Differential Equation
  • Linear Differential Equation
  • Separable Differential Equation
  • Homogeneous Differential Equation
  • Integrate by partial fractions
  • Product of Binomials with Common Term
  • FOIL Method
  • Load more...
Can't find a method? Tell us so we can add it.

I. Express the LHS in terms of sine and cosine and simplify

1

Start from the LHS (left-hand side)

$\sec\left(x\right)^2+\csc\left(x\right)^2$
2

Rewrite $\sec\left(x\right)$ in terms of sine and cosine

$\left(\frac{1}{\cos\left(x\right)}\right)^2+\csc\left(x\right)^2$
3

The power of a quotient is equal to the quotient of the power of the numerator and denominator: $\displaystyle\left(\frac{a}{b}\right)^n=\frac{a^n}{b^n}$

$\frac{1}{\cos\left(x\right)^2}+\csc\left(x\right)^2$
4

Applying the cosecant identity: $\displaystyle\csc\left(\theta\right)=\frac{1}{\sin\left(\theta\right)}$

$\frac{1}{\cos\left(x\right)^2}+\frac{1}{\sin\left(x\right)^2}$
5

Combine fractions with different denominator using the formula: $\displaystyle\frac{a}{b}+\frac{c}{d}=\frac{a\cdot d + b\cdot c}{b\cdot d}$

$\frac{\sin\left(x\right)^2+\cos\left(x\right)^2}{\cos\left(x\right)^2\sin\left(x\right)^2}$
6

Applying the pythagorean identity: $\sin^2\left(\theta\right)+\cos^2\left(\theta\right)=1$

$\frac{1}{\cos\left(x\right)^2\sin\left(x\right)^2}$
Why is sin(x)^2 + cos(x)^2 = 1 ?

II. Express the RHS in terms of sine and cosine and simplify

7

Start from the RHS (right-hand side)

$\frac{1}{\sin\left(x\right)^2\cos\left(x\right)^2}$
8

Multiply $\frac{1}{\sin\left(x\right)^2\cos\left(x\right)^2}$ by $\frac{sin(x)^2+cos(x)^2}{sin(x)^2+cos(x)^2}$

$\frac{1}{\sin\left(x\right)^2\cos\left(x\right)^2}\frac{\sin\left(x\right)^2+\cos\left(x\right)^2}{\sin\left(x\right)^2+\cos\left(x\right)^2}$
9

Multiplying fractions $\frac{1}{\sin\left(x\right)^2\cos\left(x\right)^2} \times \frac{\sin\left(x\right)^2+\cos\left(x\right)^2}{\sin\left(x\right)^2+\cos\left(x\right)^2}$

$\frac{\sin\left(x\right)^2+\cos\left(x\right)^2}{\sin\left(x\right)^2\cos\left(x\right)^2\left(\sin\left(x\right)^2+\cos\left(x\right)^2\right)}$
10

Applying the pythagorean identity: $\sin^2\left(\theta\right)+\cos^2\left(\theta\right)=1$

$\frac{\sin\left(x\right)^2+\cos\left(x\right)^2}{\sin\left(x\right)^2\cos\left(x\right)^2}$
Why is sin(x)^2 + cos(x)^2 = 1 ?
11

Expand the fraction $\frac{\sin\left(x\right)^2+\cos\left(x\right)^2}{\sin\left(x\right)^2\cos\left(x\right)^2}$ into $2$ simpler fractions with common denominator $\sin\left(x\right)^2\cos\left(x\right)^2$

$\frac{\sin\left(x\right)^2}{\sin\left(x\right)^2\cos\left(x\right)^2}+\frac{\cos\left(x\right)^2}{\sin\left(x\right)^2\cos\left(x\right)^2}$
12

Simplify the resulting fractions

$\frac{1}{\cos\left(x\right)^2}+\frac{1}{\sin\left(x\right)^2}$
13

Since $\cos$ is the reciprocal of $\sec$, $\frac{1}{\cos\left(x\right)^2}$ is equivalent to $\sec\left(x\right)^2$

$\sec\left(x\right)^2+\frac{1}{\sin\left(x\right)^2}$
14

Since $\sin$ is the reciprocal of $\csc$, $\frac{1}{\sin\left(x\right)^2}$ is equivalent to $\csc\left(x\right)^2$

$\sec\left(x\right)^2+\csc\left(x\right)^2$
15

Rewrite $\sec\left(x\right)$ in terms of sine and cosine

$\left(\frac{1}{\cos\left(x\right)}\right)^2+\csc\left(x\right)^2$
16

Applying the cosecant identity: $\displaystyle\csc\left(\theta\right)=\frac{1}{\sin\left(\theta\right)}$

$\left(\frac{1}{\cos\left(x\right)}\right)^2+\frac{1}{\sin\left(x\right)^2}$
17

Combine all terms into a single fraction with $\sin\left(x\right)^2$ as common denominator

$\frac{\left(\frac{1}{\cos\left(x\right)}\right)^2\sin\left(x\right)^2+1}{\sin\left(x\right)^2}$
18

Applying the trigonometric identity: $\sin\left(\theta \right)^2 = 1-\cos\left(\theta \right)^2$

$\frac{\left(\frac{1}{\cos\left(x\right)}\right)^2\sin\left(x\right)^2+1}{1-\cos\left(x\right)^2}$
Why is 1 - cos(x)^2 = sin(x)^2 ?
19

The power of a quotient is equal to the quotient of the power of the numerator and denominator: $\displaystyle\left(\frac{a}{b}\right)^n=\frac{a^n}{b^n}$

$\frac{\frac{1}{\cos\left(x\right)^2}\sin\left(x\right)^2+1}{1-\cos\left(x\right)^2}$
20

Multiply the fraction by the term

$\frac{\frac{\sin\left(x\right)^2}{\cos\left(x\right)^2}+1}{1-\cos\left(x\right)^2}$
21

Simplify the fraction $\frac{\sin\left(x\right)^2}{\cos\left(x\right)^2}+1$ into $\frac{1}{\cos\left(x\right)^2}$

$\frac{\frac{1}{\cos\left(x\right)^2}}{1-\cos\left(x\right)^2}$
22

Divide fractions $\frac{\frac{1}{\cos\left(x\right)^2}}{1-\cos\left(x\right)^2}$ with Keep, Change, Flip: $\frac{a}{b}\div c=\frac{a}{b}\div\frac{c}{1}=\frac{a}{b}\times\frac{1}{c}=\frac{a}{b\cdot c}$

$\frac{1}{\cos\left(x\right)^2\left(1-\cos\left(x\right)^2\right)}$
23

Multiply the single term $\cos\left(x\right)^2$ by each term of the polynomial $\left(1-\cos\left(x\right)^2\right)$

$\frac{1}{\cos\left(x\right)^2-\cos\left(x\right)^{4}}$

III. Choose what side of the identity are we going to work on

24

To prove an identity, we usually begin to work on the side of the equality that seems to be more complicated, or the side that is not expressed in terms of sine and cosine. In this problem, we will choose to work on the left side $\frac{1}{\cos\left(x\right)^2\sin\left(x\right)^2}$ to reach the right side $\frac{1}{\cos\left(x\right)^2-\cos\left(x\right)^{4}}$

$\frac{1}{\cos\left(x\right)^2\sin\left(x\right)^2}=\frac{1}{\cos\left(x\right)^2-\cos\left(x\right)^{4}}$
25

Rewrite $\frac{1}{\cos\left(x\right)^2\sin\left(x\right)^2}$ as $\sec\left(x\right)^2+\csc\left(x\right)^2$ by applying trigonometric identities

$\sec\left(x\right)^2+\csc\left(x\right)^2$
26

Rewrite $\sec\left(x\right)$ in terms of sine and cosine

$\left(\frac{1}{\cos\left(x\right)}\right)^2+\csc\left(x\right)^2$
27

Applying the cosecant identity: $\displaystyle\csc\left(\theta\right)=\frac{1}{\sin\left(\theta\right)}$

$\left(\frac{1}{\cos\left(x\right)}\right)^2+\frac{1}{\sin\left(x\right)^2}$
28

Combine all terms into a single fraction with $\sin\left(x\right)^2$ as common denominator

$\frac{\left(\frac{1}{\cos\left(x\right)}\right)^2\sin\left(x\right)^2+1}{\sin\left(x\right)^2}$
29

Applying the trigonometric identity: $\sin\left(\theta \right)^2 = 1-\cos\left(\theta \right)^2$

$\frac{\left(\frac{1}{\cos\left(x\right)}\right)^2\sin\left(x\right)^2+1}{1-\cos\left(x\right)^2}$
Why is 1 - cos(x)^2 = sin(x)^2 ?
30

The power of a quotient is equal to the quotient of the power of the numerator and denominator: $\displaystyle\left(\frac{a}{b}\right)^n=\frac{a^n}{b^n}$

$\frac{\frac{1}{\cos\left(x\right)^2}\sin\left(x\right)^2+1}{1-\cos\left(x\right)^2}$
31

Multiply the fraction by the term

$\frac{\frac{\sin\left(x\right)^2}{\cos\left(x\right)^2}+1}{1-\cos\left(x\right)^2}$
32

Simplify the fraction $\frac{\sin\left(x\right)^2}{\cos\left(x\right)^2}+1$ into $\frac{1}{\cos\left(x\right)^2}$

$\frac{\frac{1}{\cos\left(x\right)^2}}{1-\cos\left(x\right)^2}$
33

Divide fractions $\frac{\frac{1}{\cos\left(x\right)^2}}{1-\cos\left(x\right)^2}$ with Keep, Change, Flip: $\frac{a}{b}\div c=\frac{a}{b}\div\frac{c}{1}=\frac{a}{b}\times\frac{1}{c}=\frac{a}{b\cdot c}$

$\frac{1}{\cos\left(x\right)^2\left(1-\cos\left(x\right)^2\right)}$
34

Multiply the single term $\cos\left(x\right)^2$ by each term of the polynomial $\left(1-\cos\left(x\right)^2\right)$

$\frac{1}{\cos\left(x\right)^2-\cos\left(x\right)^{4}}$

IV. Check if we arrived at the expression we wanted to prove

35

Since we have reached the expression of our goal, we have proven the identity

true

Final answer to the problem

true

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Similar Problems Solved

Prove the identity

$\frac{1}{1-sin\left(x\right)}-\frac{1}{1+\sin\left(x\right)}=2\tan\left(x\right)sec\left(x\right)$

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$\frac{1+\cos2x}{\sin2x}=\cot x$

Prove the identity

$sec\left(x\right)-tan\left(x\right)sin\left(x\right)=\frac{1}{sec\left(x\right)}$

Prove the identity

$\tan\left(x\right)+\cot\left(x\right)=\sec\left(x\right)\csc\left(x\right)$

Prove the identity

$\tan\left(a\right)+\cot\left(a\right)=\sec\left(a\right)\csc\left(a\right)$

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$\sec x-\sin x\tan x=\cos x$

Main Topic: Trigonometric Identities

In mathematics, trigonometric identities are equalities that involve trigonometric functions and are true for every single value of the occurring variables where both sides of the equality are defined.

Used Formulas

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