If sin A=2/3, cos B=1/3, find (sec A)^2-(tan B)^2.

Answer 1

use a calculator with a trig functions

sinA = 2/3

A = 41.81 degrees

cosB = 1/3

B = 70.53

secA = sec41.81 = 1/cos41.81 = 1.34

sec^2(A) = 1.34^2 = 1.8

tanB = tan70.53 = 2.828

tan^2(B) = 2.828^2 = 8

sec^2(A) -tan^2(B) = 1.8-8 = -6.2

or calculate the adjacent and opposite sides with the Pythagorean Theorem

sinA = 2/3 = opposite side/hypotenuse

cosA = adjacent side/hypotenuse =sqr(9-4)/3 = sqr5/3

secA = 1/cosA = 3/sqr5

sec^2(A) = 9/5 = 1.8

cosB = 1/3

sinB = sqr(9-1)/3 = sqr8/3

tanB = opposite side/adjacent side= sinB/cosB = (sqr8/3)/(1/3) = sqr8

tan^2(B) = 8

sec^2(A) - tan^2(B) = 1.8-8 = -6.2

Answer 2

Using the formula sin²A + cos²A=1, we can find the value of cosA

cosA = √(1-sin²A)

=√(1-(2/3)²)

= √1-(4/9)

= √(5/9)

= √(5)/3

SecA = 1/cosA (Cos and Sec are reciprocal of each other)

=1/(√(5)/3)

=3/√5

Sec²A = 9/5

CosB = 1/3. Using similar formula and idea above, we can find sinB = √8/3

TanB = SinB/CosB

= (√8/3)/(1/3)

=√8

Tan²B = 8

Hence, Sec²A - Tan²B = 9/5 - 8

= 1.8-8

= -6.2