tan^2t/sect=sect-cost
cott^2/csct=csct-sint
Are you showing that they are identities ?
if so,
LS = (sin^2t/cos^2t)(cost)
= sin^2t/cost
RS = 1/cost - cost
= (1 - cos^2t)/cost
= sin^2t/cost
= LS
The second one works the same way, try it.
To simplify the given trigonometric equations, let's break them down step by step:
1. tan^2(t) / sec(t) = sec(t) - cos(t)
Let's manipulate both sides of the equation separately to simplify the expression.
On the left side:
tan^2(t) = sin^2(t) / cos^2(t) (using the identity tan(t) = sin(t) / cos(t))
Now, substituting this value in the equation:
sin^2(t) / cos^2(t) / sec(t) = sec(t) - cos(t)
Next, we can simplify the left side:
sin^2(t) * cos(t) / cos^2(t) = sec(t) - cos(t)
sin^2(t) * cos(t) = sec(t) * cos^2(t) - cos^3(t)
sin^2(t) * cos(t) = cos(t) - cos^3(t)
From here, we can multiply both sides by cos(t) to eliminate the denominator:
sin^2(t) * cos^2(t) = cos(t) * cos^2(t) - cos^3(t)
sin^2(t) * cos^2(t) = cos^3(t) - cos^3(t)
sin^2(t) * cos^2(t) = 0
Therefore, the equation simplifies to sin^2(t) * cos^2(t) = 0.
2. cot^2(t) / csc(t) = csc(t) - sin(t)
To simplify this equation, we'll follow similar steps as above.
On the left side:
cot^2(t) = cos^2(t) / sin^2(t) (using the identity cot(t) = cos(t) / sin(t))
Now, substituting this value in the equation:
cos^2(t) / sin^2(t) / csc(t) = csc(t) - sin(t)
Next, we can simplify the left side:
cos^2(t) * sin(t) / sin^2(t) = csc(t) - sin(t)
cos^2(t) * sin(t) = sin(t) * sin^2(t) - cos(t) * sin^2(t)
cos^2(t) * sin(t) = sin(t) - cos(t) * sin^2(t)
From here, we need to use the identity sin^2(t) = 1 - cos^2(t):
cos^2(t) * sin(t) = sin(t) - cos(t) * (1 - cos^2(t))
cos^2(t) * sin(t) = sin(t) - cos(t) + cos^3(t)
Now, multiplying both sides by sin(t) to eliminate the denominator:
cos^2(t) * sin^2(t) = sin(t) * sin(t) - cos(t) * sin(t) + cos^3(t)
cos^2(t) * sin^2(t) = sin^2(t) - sin(t) * cos(t) + cos^3(t)
Therefore, the equation simplifies to cos^2(t) * sin^2(t) = sin^2(t) - sin(t) * cos(t) + cos^3(t).
Remember to double-check the steps and equations for accuracy.