verify following trigonometric identities. 1/sec0tan0 = csc0 = sin0
check your typing, your statement is not true.
I figured you must mean
1/(secØtanØ) - cscØ = - sinØ
Proof:
LS = 1/((1/cosØ)(sinØ/cosØ)) - 1/sinØ
= cos^2 Ø/sinØ - 1/sinØ
= (cos^ Ø - 1)sinØ
= - sin^2 Ø/sinØ
= =sinØ
= RS
To verify the trigonometric identity 1/sec(θ)tan(θ) = csc(θ) = sin(θ), we need to demonstrate that both sides of the equation are equal for all valid values of θ.
Let's start by simplifying each side of the equation:
1/sec(θ)tan(θ)
To simplify this expression, we will use reciprocal trigonometric identities:
sec(θ) = 1/cos(θ)
tan(θ) = sin(θ)/cos(θ)
So, 1/sec(θ)tan(θ) becomes:
1 / (1/cos(θ)) * (sin(θ)/cos(θ))
By multiplying the numerator and denominator by cos(θ), we can simplify this expression further:
1 / (1/cos(θ)) * (sin(θ)/cos(θ))
= cos(θ) * sin(θ)
= sin(θ)cos(θ)
Now, let's simplify the right side of the equation:
csc(θ) = 1/sin(θ)
Finally, we compare both sides of the equation:
sin(θ)cos(θ) = 1/sin(θ)
To simplify the right side, we multiply the numerator and denominator by sin(θ):
sin(θ)cos(θ) = (1 * sin(θ)) / (sin(θ) * sin(θ))
This simplifies to:
sin(θ)cos(θ) = sin(θ) / sin(θ)^2
Using the identity sin(θ)^2 + cos(θ)^2 = 1, we can rewrite the right side of the equation as:
sin(θ)cos(θ) = sin(θ) / (1 - cos(θ)^2)
Since cos(θ)^2 is equal to 1 - sin(θ)^2, we can substitute this value:
sin(θ)cos(θ) = sin(θ) / (1 - (1 - sin(θ)^2))
= sin(θ) / (sin(θ)^2)
Now, we can see that the right side of the equation simplifies to:
sin(θ)cos(θ) = 1/sin(θ)
As we can see, both sides of the equation are equal, which verifies the trigonometric identity 1/sec(θ)tan(θ) = csc(θ) = sin(θ) for all valid values of θ.