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

Question

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

Answer 1

use a calculator with inverse trig functions

sinA = 2/3, A= sin^-1(2/3) = 41.81 degrees

cosB = 1/3, B = cos^-1(1/3) = 70.53 degrees

tanB = tan70.53 = 2.8284. 2.8284^2 = 8

secA = 1/cosA = 1/cos41.81 =1.3416. 1.3416^2 = 1.8

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

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