tan(7π/4) + tan(5π/4)
Does this = 0
Tan7pi/4=Tan(pi+3pi/4) lies in IVQdt and is =-Tan pi/4.
Tan 5pi/4=tan (pi+pi/4) lis in IIIQdt and is=tan pi/4, hence your ans is correct.
Tan (7pi/4)+ Tan (5pi/4)
=Tan (pi+3pi/4)+ Tan (pi+pi/4)
=-Tan (pi/4)+ Tan (pi/4)
= 0
To determine if the expression tan(7π/4) + tan(5π/4) is equal to 0, you can use the properties and identities of the tangent function.
First, let's simplify each term individually:
tan(7π/4):
To simplify tan(7π/4), we need to determine the angle in the first full revolution that is equivalent to 7π/4. Since one full revolution is equal to 2π, we can subtract 2π from 7π/4 to find the equivalent angle.
7π/4 - 2π = (7π - 8π) / 4 = -π/4
Therefore, tan(7π/4) is equivalent to tan(-π/4). Since the tangent function has a period of π, we can find an equivalent angle in the first full revolution by adding π to -π/4.
-π/4 + π = (4π - π) / 4 = 3π/4
So, tan(7π/4) is equivalent to tan(3π/4).
Similarly, let's simplify tan(5π/4):
5π/4 - 2π = (5π - 8π) / 4 = -3π/4
Therefore, tan(5π/4) is equivalent to tan(-3π/4).
Now, we can calculate the values of tan(3π/4) and tan(-3π/4). The tangent function is positive in the first and third quadrants, so tan(3π/4) is equal to tan(-3π/4).
Next, we can use the identity that tan(-θ) = -tan(θ). So, tan(-3π/4) = -tan(3π/4).
Since tan(3π/4) and tan(-3π/4) are negative reciprocals of each other, their sum will result in 0:
tan(3π/4) + tan(-3π/4) = 0
Therefore, the expression tan(7π/4) + tan(5π/4) does equal 0.