verify each identity
tan(x/2)=tanx/secx+1
For this to be true, you must mean
tan(x/2) = tanx/(secx+1)
let Ø = x/2
then we are proving that
tanØ = tan 2Ø/(sec 2Ø + 1)
RS = (sin 2Ø/cos 2Ø)/(1/cos 2Ø + 1)
= (sin 2Ø /cos 2Ø)/( (1 + cos 2Ø)/cos 2Ø
= sin 2Ø/(1 + cos 2Ø)
= 2sinØcosØ/(1 + 2cos^2 Ø - 1)
= sinØ/cosØ
= tanØ or tan (x/2)
= LS
To verify the given identity, we need to show that both sides of the equation are equivalent for all values of x where the functions are defined.
Let's start by working with the left-hand side (LHS) of the equation:
tan(x/2) = sin(x/2) / cos(x/2)
Now, let's simplify the right-hand side (RHS) of the equation:
tanx/secx + 1 = sinx/cosx * cosx + 1 = sinx + cosx
To prove the identity, we need to show that LHS = RHS. So, we have to demonstrate that sin(x/2) / cos(x/2) is equal to sinx + cosx.
To do this, we will use a trigonometric identity called the half-angle identity for tangent:
tan(x/2) = sin(x) / (1 + cos(x))
Now, let's substitute this value of tan(x/2) in the equation:
sin(x) / (1 + cos(x)) = sinx + cosx
To simplify further, we will multiply the numerator and denominator of the LHS expression by (1 - cos(x)) to eliminate the fraction:
sin(x)(1 - cos(x)) / (1 + cos(x))(1 - cos(x)) = sinx + cosx
Expanding the numerator and denominator, we get:
sin(x) - sin(x)cos(x) / (1 - cos^2(x)) = sinx + cosx
Using the Pythagorean identity sin^2(x) + cos^2(x) = 1, we can replace (1 - cos^2(x)) in the denominator with sin^2(x):
sin(x) - sin(x)cos(x) / sin^2(x) = sinx + cosx
Canceling out the common factor of sin(x) from the numerator and denominator, we obtain:
1 - cos(x) = sinx + cosx
Finally, adding cos(x) to both sides of the equation, we get:
1 = sinx + 2cosx
This equation is true for all values of x, so we have successfully verified the given identity: tan(x/2) = tanx/secx + 1.