How do you find sec 2x, given tan x = 5/3 and sin x< 0?

1 Answer
Feb 9, 2016

Answer:

#-34/16#

Explanation:

Starting from tan x, you can find sec x, because of the trigonometric identity 1 + #tan^2x# = # sec^2x#
1+#(5/3)^2# =# sec^2x#
# secx#= #sqrt(34/9)#

But since x is in Quadrant II, sec x has to be negative. That's because sec x has the same sign as cos x, because sec x = 1 / cos x. We know that cos x is negative is Quadrant II, therefore so is sec x. So,# secx#= - #sqrt(34/9)# = - #sqrt(34)/3#

Since sec x and cos x are reciprocals of each other,
cos x = 1/sec x = - #3/sqrt(34)#
#cos^2x=# 9/34.......... eq (i)

Now use the identity #sin^2x+cos^2x=1# to find sin x:
#sin^2x#= #1-(9/34)# =25/34 ......eq (ii)
Again, we know that sin x is positive in Quadrant II

We know,
sec2x= #1/cos(2x)#
= #1/(cos^2x-Sin^2x)#
substituting the values,
sec2x= #1/((9/34)-(25/34)#= #-34/16#