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#