Verify the trigonometric identity:
[(1–sin²x)/sin²x]–[(csc²x–1)/cos²x]=
-tan²x
I can't figure this out.
To verify the trigonometric identity, we'll start by simplifying each side of the equation.
Left side of the equation:
[(1–sin²x)/sin²x]–[(csc²x–1)/cos²x]
We can simplify the first term by using the identity sin²x = 1 - cos²x, which gives us:
[(1 - (1 - cos²x))/sin²x] - [(csc²x - 1)/cos²x]
Simplifying further, we have:
[(cos²x)/sin²x] - [(csc²x - 1)/cos²x]
Next, we'll find a common denominator for both terms, which is cos²x * sin²x:
[(cos²x * cos²x)/ (sin²x * cos²x)] - [(csc²x - 1)(sin²x * sin²x)/(cos²x * sin²x)]
This simplifies to:
[(cos⁴x)/(sin²x * cos²x)] - [(csc²x * sin⁴x - sin⁴x)/(cos²x * sin²x)]
Now, we can simplify further using the reciprocal identities csc²x = 1/sin²x and cos²x = 1 - sin²x:
[(cos⁴x)/(sin²x * (1 - sin²x))] - [(1/sin²x * sin⁴x - sin⁴x)/(1 - sin²x)]
Simplifying the numerators, we have:
[(cos⁴x)/(sin²x - sin⁴x)] - [(1 - sin²x * sin⁴x - sin⁴x)/(1 - sin²x)]
Combining like terms, we get:
[(cos⁴x)/(sin²x - sin⁴x)] - [(1 - 2sin⁴x)/(1 - sin²x)]
Using the identity sin²x = 1 - cos²x, we can simplify further:
[(cos⁴x)/(sin²x - sin⁴x)] - [(1 - 2sin⁴x)/(cos²x)]
Factoring out sin²x from the denominator of the first term, we have:
[(cos⁴x)/(sin²x(1 - sin²x))] - [(1 - 2sin⁴x)/(cos²x)]
Simplifying the denominators, we get:
[(cos⁴x)/(sin²x * cos²x)] - [(1 - 2sin⁴x)/(cos²x)]
Now, we can combine the two terms using a common denominator, which is sin²x * cos²x:
[(cos⁴x - (1 - 2sin⁴x))/(sin²x * cos²x)]
Combining like terms in the numerator, we get:
[(cos⁴x - 1 + 2sin⁴x)/(sin²x * cos²x)]
Using the identity cos²x = 1 - sin²x, we can simplify further:
[(1 - sin²x)² - 1 + 2sin⁴x)/(sin²x * (1 - sin²x))]
Expanding and simplifying (1 - sin²x)² - 1, we get:
[(1 - 2sin²x + sin⁴x - 1 + 2sin⁴x)/(sin²x * (1 - sin²x))]
The numerator cancels out:
(0)/(sin²x * (1 - sin²x))
This simplifies to:
0
Right side of the equation:
-tan²x
Since 0 is equal to 0, we can conclude that the trigonometric identity is verified:
[(1–sin²x)/sin²x]–[(csc²x–1)/cos²x] = -tan²x
To verify the given trigonometric identity, we need to simplify both sides of the equation and show that they are equal.
Starting from the left side of the equation:
[(1–sin²x)/sin²x] – [(csc²x–1)/cos²x]
First, let's simplify the numerator and denominator separately.
Numerator:
1 – sin²x
Using the identity sin²x + cos²x = 1, we can rewrite this as:
cos²x – sin²x
Denominator:
sin²x
Using the reciprocal identity csc x = 1/sin x, we can rewrite this as:
(1/csc²x)
Substituting the simplified forms into the given equation, we get:
[(cos²x – sin²x)/(1/csc²x)] – [(csc²x – 1)/cos²x]
Next, we'll simplify each term individually:
Term 1: [(cos²x – sin²x)/(1/csc²x)]
To simplify this fraction, divide the numerator by the reciprocal of the denominator:
(cos²x – sin²x) * csc²x
Using the Pythagorean Identity sin²x = 1 – cos²x, we can rewrite this as:
(cos²x – (1 – cos²x)) * csc²x
(cos²x – 1 + cos²x) * csc²x
(2cos²x – 1) * csc²x
Term 2: [(csc²x – 1)/cos²x]
To simplify this fraction, divide the numerator by the denominator:
(csc²x – 1) / cos²x
Using the reciprocal identity csc x = 1/sin x, we can rewrite this as:
[(1/sin²x) – 1]/cos²x
Now, let's combine both terms together:
[(2cos²x – 1) * csc²x] – [(1/sin²x) – 1]/cos²x
To simplify further, let's work with each term individually:
Term 1: (2cos²x – 1) * csc²x
This can be rewritten as:
[(2cos²x – 1)/sin²x]
Term 2: [(1/sin²x) – 1]/cos²x
To simplify this fraction, we need to find a common denominator. The common denominator is sin²x * cos²x:
[(cos²x – sin²x)/sin²x * cos²x]
Now, let's combine both terms:
[(2cos²x – 1)/sin²x] – [(cos²x – sin²x)/sin²x * cos²x]
Next, let's simplify the numerators:
Numerator of Term 1: (2cos²x – 1)
Using the identity 1 – sin²x = cos²x, we can rewrite this as:
(2 (1 – sin²x) – 1)
(2cos²x – 2sin²x) – 1
2cos²x – 2sin²x – 1
Numerator of Term 2: (cos²x – sin²x)
Using the identity 1 – cos²x = sin²x, we can rewrite this as:
(cos²x – (1 – cos²x))
(cos²x – 1 + cos²x)
2cos²x – 1
Now, let's substitute the simplified numerators into the equation:
[(2cos²x – 2sin²x – 1)/sin²x] – [(2cos²x – 1)/(sin²x * cos²x)]
Now, let's simplify further by finding the common denominator for the two fractions, which is sin²x * cos²x:
[(2cos²x – 2sin²x – 1)/sin²x] – [(2cos²x – 1)/(sin²x * cos²x)]
To do this, we need to multiply the first fraction by cos²x/cos²x and the second fraction by sin²x/sin²x:
[(2cos²x – 2sin²x – 1)/(sin²x * cos²x)] – [(2cos²x – 1)/(sin²x * cos²x)]
After multiplying, we get:
[(2cos²x – 2sin²x – 1) – (2cos²x – 1)] / (sin²x * cos²x)
Simplifying further:
[(2cos²x – 2sin²x – 1 – 2cos²x + 1)] / (sin²x * cos²x)
Now, let's cancel out like terms:
[-2sin²x] / (sin²x * cos²x)
To simplify this further, let's use the identity sin x/cos x = tan x:
-2sin²x / (sin²x * cos²x) = -2 * (sin x / cos x) * (sin x / cos x)
= -2 * (sin x / cos x)²
= -2tan²x
Therefore, the left side of the equation simplifies to -2tan²x.
Compared to the right side of the equation, which is also -tan²x, we have proved that the given trigonometric identity is true.