If sin of theta equals 3/8 and theta is in quadrant II. what are cos, tan, csc, cot, and sec of theta?

#sintheta = 3/8#, #theta# in quadrant #II#

Imagine a right triangle being drawn on the cartesian plane, as in the following example.

Since sin = opposite/hypotenuse, the side opposite to #theta# is 3 and the hypotenuse is #8#, we can rearrange our pythagorean theorem to find the adjacent side, b.

#a^2 + b^2 = c^2#

#b^2 = c^2 - a^2#

#b^2 = 8^2 - 3^2#

#b^2 = 64 - 9#

#b^2 = 55#

#b = -sqrt(55)#

So, now we know that the adjacent side measures #-sqrt(55)# units (since the x axis is negative in quadrant II) in length. Thus we can deduce that #costheta = -sqrt(55)/8# and #tantheta = -3/(sqrt(55))#

Now, for #csctheta, sectheta and cottheta#, we must apply the reciprocal identities.

#csctheta = 1/sintheta#

#sectheta = 1/costheta#

#cottheta = 1/tantheta#

Therefore, #csctheta = 8/3, sectheta = -8/sqrt(55) and cottheta = -sqrt(55)/3#

To summarize, the six trigonometric ratios, we get:

#sintheta = 3/8#

#costheta = -sqrt(55)/8#

#tantheta = -3/sqrt(55)#

#csctheta = 8/3#

#sectheta = -8/sqrt(55)#

#cottheta = -sqrt(55)/3#

Hopefully this helps!

#sintheta=3/8to(color(red)(1))#

#• csctheta=1/sintheta=8/3to(color(red)(2))#

#• costheta=+-sqrt(1-sin^2theta)#

#color(white)(xxxxx)=-sqrt(1-9/64)larr" negative value"#

#color(white)(xxxxx)=-sqrt(55/64)#

#rArrcostheta=-sqrt55/8to(color(red)(3))#

#• sectheta=1/costheta=-8/sqrt55to(color(red)(4))#

#• tantheta=(sintheta)/(costheta)#

#color(white)(xxxx)=3/8xx-8/sqrt55=-3/sqrt55to(color(red)(5))#

#• cottheta=1/tantheta=-sqrt55/3to(color(red)(6))#