If sin(theta)=3/5 , tan(phi)= 1/2 and 90¡ã< theta <180¡ã<phi< 270¡ã, find tha value of 8tan(theta) - 5^1/2 sec(phi)
since θ is in QII
and Ø is in QIII,
tanθ = 3/-4
secØ = √5/-2
now go figure.
To find the value of 8tan(theta) - √5 sec(phi), we need to substitute the given values of sin(theta) and tan(phi) into the expression and simplify.
Given:
sin(theta) = 3/5
tan(phi) = 1/2
90° < theta < 180°
180° < phi < 270°
Step 1: Find cos(theta) using the Pythagorean identity for sine and cosine.
cos^2(theta) = 1 - sin^2(theta)
cos^2(theta) = 1 - (3/5)^2
cos^2(theta) = 1 - 9/25
cos^2(theta) = 16/25
cos(theta) = ± √(16/25)
cos(theta) = ± (4/5)
Since 90° < theta < 180°, cos(theta) is negative.
cos(theta) = -4/5
Step 2: Find sec(phi) using the reciprocal identity for cosine.
sec(phi) = 1/cos(phi)
sec(phi) = 1/√(1 - sin^2(phi))
sec(phi) = 1/√(1 - (1/2)^2)
sec(phi) = 1/√(1 - 1/4)
sec(phi) = 1/√(3/4)
sec(phi) = 1/√(3/4) * √(4/4)
sec(phi) = 1/√(3/16)
sec(phi) = 1/(√3/4)
sec(phi) = 4/(√3)
sec(phi) = (4√3) / 3
Step 3: Substitute the values into the expression and simplify.
8tan(theta) - √5 sec(phi) = 8 * (sin(theta) / cos(theta)) - √5 * (4√3) / 3
Substituting sin(theta) and cos(theta):
= 8 * (3/5) / (-4/5) - √5 * (4√3) / 3
Simplifying:
= -24/4 - (4√3√5) / 3
= -6 - (4√3√5) / 3
Therefore, the value of 8tan(theta) - √5 sec(phi) is -6 - (4√3√5) / 3.