If sinA =3/5,then what is the value of cot²A+sec²A ?

1 Answer
Feb 25, 2018

cot^2(A)+sec^2(A)=57/16

Explanation:

Recall that cot(x)=cos(x)/sin(x) and sec(x)=1/cos(x).

cot(A)=cos(A)/sin(A)

cot^2(A)=cos^2(A)/sin^2(A)

sec(A)=1/cos(A)

sec^2(A)=1/cos^2(A)

If sin(A)=3/5, sin^2(A)=(3/5)^2=9/25

We have sin^2(A); however, we still need cos^2(A) to determine the values of cot^2(A) and sec^2(A).

Recall the following identity:

sin^2(x)+cos^2(x)=1

So,

sin^2(A)+cos^2(A)=1

cos^2(A)=1-sin^2(A)

We know the value of sin^2(A), so

cos^2(A)=1-9/25=25/25-9/25=16/25

cot^2(A)=cos^2(A)/sin^2(A)=(16/25)/(9/25)=16/cancel25 * cancel25/9=16/9 (Because (a/b)/(c/d)=a/b*d/c)

sec^2(A)=1/cos^2(A)=1/(16/25)=25/16 (Because 1/(a/b)=b/a).

cot^2(A)+sec^2(A)=16/9+25/16=32/16+25/16=57/16