Prove the following trigonometric identities
1. tanx = sinx + sin^2x/cosx(1+sinx)
2. cos^3+(cosx)(sin^2x) = 1/secx
1. The way you typed it, the identity is false, but I am sure you meant
tanx = (sinx + sin^2x)/(cosx(1 + sinx))
RS = sinx(1 + sinx) / (cosx(1+sinx))
= sinx/cosx
= tanx
= LS
2. LS = cosx(cos^2x + sin^2x)
= (cosx)(1)
= cosx
RS = 1/secx
= cosx
= LS
To prove trigonometric identities, we need to manipulate the expressions using basic trigonometric relationships and algebraic techniques. Let's prove each of the given identities step by step.
1. Proving tan(x) = (sin(x) + sin^2(x)) / (cos(x) * (1 + sin(x)))
Starting with the left-hand side (LHS):
LHS = tan(x)
Using the definition of tangent:
LHS = sin(x) / cos(x)
Now, let's work on the right-hand side (RHS):
RHS = (sin(x) + sin^2(x)) / (cos(x) * (1 + sin(x)))
Using the identity sin^2(x) = 1 - cos^2(x):
RHS = (sin(x) + (1 - cos^2(x))) / (cos(x) * (1 + sin(x)))
Simplifying the numerator:
RHS = (1 + sin(x) - cos^2(x)) / (cos(x) * (1 + sin(x)))
Factoring out cos(x) from the numerator:
RHS = (1 + sin(x) - cos^2(x)) / (cos(x) * (1 + sin(x)))
= (1 - cos^2(x) + sin(x)) / (cos(x) * (1 + sin(x)))
Using the identity 1 - cos^2(x) = sin^2(x):
RHS = (sin^2(x) + sin(x)) / (cos(x) * (1 + sin(x)))
Dividing both the numerator and denominator by cos(x):
RHS = (sin(x) + sin^2(x)) / (cos(x) * (1 + sin(x)))
Therefore, LHS = RHS, which verifies the given trigonometric identity.
2. Proving cos^3(x) + (cos(x) * sin^2(x)) = 1 / sec(x)
Starting with the left-hand side (LHS):
LHS = cos^3(x) + (cos(x) * sin^2(x))
Using the identity sec^2(x) = 1 + tan^2(x):
LHS = cos^3(x) + (cos(x) * sin^2(x))
= cos(x) * (cos^2(x) + sin^2(x))
= cos(x) * 1
= cos(x)
Now, let's work on the right-hand side (RHS):
RHS = 1 / sec(x)
Using the identity sec(x) = 1 / cos(x):
RHS = 1 / (1 / cos(x))
= cos(x)
Therefore, LHS = RHS, which verifies the given trigonometric identity.