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Question

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

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Solution

We have,

cosec2θcot2θ=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, sec2θtan2θ=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=11312=1213

and, sinθ=1cosecθ=1135=513

Hence, sinθ=513,cosθ=1213,

tanθ=512,secθ=1312 and cosecθ

=135

(ii)We have,

sin2θ+cos2θ=1

sin2θ=1cos2θ

sinθ=±1cos2θ

In the 2nd quadrant sinθ is positive and tanθ is negative.

sinθ=1cos2θ

=1(12)2[cosθ=12]=114

=34=32

and, tanθ=sinθcosθ=3212=3

Now, cosecθ=1sinθ=132=23

secθ=1cosθ=112=2

and cotθ=1tanθ=13=13

Hence, sinθ=32,tanθ=3,

cosecθ=23,secθ=2andcotθ=13

(iii)In the third quadrant cosec θ is negative

cosecθ=1+cot2θ

=1+(43)2

=1+169

=259

=53

Now, sinθ=1cosecθ=153=35

and, cosθ=1secθ=154=45

Hence, sinθ=35,cosθ=45,

cosecθ=53,secθ=54andcotθ=43

(iv)We have,

sin2θ+cos2θ=1

cos2θ=1sin2θ

cosθ=±1sin2θ

In the 1st quadrant cosθ is positive and tan θ is also positive.

cosθ=1sin2θ

=1(35)2[sinθ=35]

=1925

=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


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