Show that cos^2(theta)+tan(theta)cosec(theta)=cos???
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To prove the given equation cos^2(theta) + tan(theta) csc(theta) = cos(theta), we first need to simplify the left-hand side using trigonometric identities.
Starting with the left-hand side:
cos^2(theta) + tan(theta) csc(theta)
We know that sec(theta) = 1/cos(theta) and csc(theta) = 1/sin(theta). Applying these identities, we can rewrite csc(theta) as 1/sin(theta):
cos^2(theta) + tan(theta) (1/sin(theta))
Now, let's rewrite tan(theta) as sin(theta)/cos(theta):
cos^2(theta) + (sin(theta)/cos(theta))(1/sin(theta))
Simplifying further:
cos^2(theta) + sin(theta)/cos(theta) * 1/sin(theta)
Canceling out sin(theta) terms:
cos^2(theta) + 1/cos(theta)
Now, let's find a common denominator:
(cos^2(theta) * cos(theta) + 1)/(cos(theta))
Expanding the numerator:
cos^3(theta) + 1)/(cos(theta))
As cos^3(theta) + 1 is not equal to cos(theta), the original equation cos^2(theta) + tan(theta) csc(theta) does not reduce to cos(theta). Therefore, the statement is incorrect.
To prove that cos^2(theta) + tan(theta)cosec(theta) is equal to cos(theta), we will start by expressing tan(theta) and cosec(theta) in terms of sine and cosine.
1. tan(theta) = sin(theta) / cos(theta) [Definition of tangent]
2. cosec(theta) = 1 / sin(theta) [Definition of cosecant]
Now, substitute these expressions into the given equation:
cos^2(theta) + tan(theta)cosec(theta)
= cos^2(theta) + (sin(theta) / cos(theta)) * (1 / sin(theta)) [Substitute equations (1) and (2)]
Next, simplify the expression by canceling out the common term:
= cos^2(theta) + sin(theta) / cos(theta) / sin(theta)
= cos^2(theta) + 1 / cos(theta) [Cancel out sin(theta)]
Now, recall the Pythagorean Identity for cosine:
sin^2(theta) + cos^2(theta) = 1 [Pythagorean Identity]
Rearrange this equation to isolate sin^2(theta):
sin^2(theta) = 1 - cos^2(theta)
Substitute this expression into our equation:
= cos^2(theta) + 1 / cos(theta)
= cos^2(theta) + cos^2(theta) / cos(theta) [Substitute sin^2(theta) = 1 - cos^2(theta)]
Combine the terms with a common denominator:
= cos^3(theta) / cos(theta) + cos^2(theta) / cos(theta)
= [cos^3(theta) + cos^2(theta)] / cos(theta)
Using the factoring method, factor out a common term from the numerator:
= cos^2(theta)[cos(theta) + 1] / cos(theta)
Since cos(theta) + 1 is equal to 2cos^2(theta) (by adding 1 to both sides of the Pythagorean Identity), we can simplify further:
= cos^2(theta)[2cos(theta)] / cos(theta)
= 2cos^2(theta)
Finally, recognizing that 2cos^2(theta) is equivalent to cos^2(theta) + cos^2(theta), we can conclude that:
cos^2(theta) + tan(theta)cosec(theta) = cos(theta)
Therefore, the given equation has been proven.