Find the values of the other five trigonometric functions in each of the following:
(i)cosθ=125,θ in quadrant III
(ii)cosθ=−12,θ in quadrant II
(iii)tanθ=34,θ in quadrant III
(iv)sinθ=35,θ in quadrant I
We have,
cosec2θ−cot2θ=1
⇒cosec2θ=1+cot2θ
⇒cosecθ=±√1+cot2θ
In the third quadrant cosec θ is negative.
∴cosec∴=−√1+cot2θ
=−√1+(125)2[∵cotθ=125]
=−√1+14425
=−√16925
=−135
∴cosecθ=−135
Now, tanθ=1cotθ
=1125=125
and, sec2θ−tan2θ=1
⇒sec2θ=1+tan2θ
secθ=±√1+tan2θ
In the third quadrant sec θ is negative
secθ=−√1+tan2θ
=−√1+(512)2[∵tanθ=512]
=−√1+25144
=−√169144=−1312
Now, cosθ=1secθ=1−1312=−1213
and, sinθ=1cosecθ=1−135=−513
Hence, sinθ=−513,cosθ=−1213,
tanθ=512,secθ=−1312 and cosecθ
=−135
(ii)We have,
sin2θ+cos2θ=1
⇒sin2θ=1−cos2θ
⇒sinθ=±√1−cos2θ
In the 2nd quadrant sinθ is positive and tanθ is negative.
∴sinθ=√1−cos2θ
=√1−(−12)2[∵cosθ=−12]=√1−14
=√34=√32
and, tanθ=sinθcosθ=√32−12=−√3
Now, cosecθ=1sinθ=1√32=2√3
secθ=1cosθ=11−2=−2
and cotθ=1tanθ=1−√3=−1√3
Hence, sinθ=√32,tanθ=−√3,
cosecθ=2√3,secθ=−2andcotθ=−1√3
(iii)In the third quadrant cosec θ is negative
∴cosecθ=−√1+cot2θ
=−√1+(43)2
=−√1+169
=−√259
=−53
Now, sinθ=1cosecθ=1−53=−35
and, cosθ=1secθ=1−54=−45
Hence, sinθ=−35,cosθ=−45,
cosecθ=−53,secθ=−54andcotθ=43
(iv)We have,
sin2θ+cos2θ=1
⇒cos2θ=1−sin2θ
⇒cosθ=±√1−sin2θ
In the 1st quadrant cosθ is positive and tan θ is also positive.
∴cosθ=√1−sin2θ
=√1−(35)2[∵sinθ=35]
=√1−925
=√1625
=45
and, tanθ=sinθcosθ=3545=34
Now, cosecθ=1sinθ=135=53
secθ=1cosθ=145=54
and, cotθ=1tanθ=134=43
Hence, cosθ=45,cosecθ=53,
tanθ=34,
secθ=54, and cotθ=43