Question

If sin theta sec theta =−1 and theta lies in the second quadrant, find sin theta and sec theta​

Answer 1

Given that :- $sin$ θ $sec$ θ $= -1$

and θ lies in the second quadrant,

As we have to find $sin$θ and $sec$θ

Therefore,

⇒ $sin^{2}$θ $=cos^{2}$θ

⇒ $sin^{2}$θ $=1-sin^{2}$θ

⇒  $sin^{2}$θ = $\frac{1}{\sqrt{2} }$

⇒ $sin$θ = ±$\frac{1}{\sqrt{2} }$

As, the θ lies in the quadrant second,

⇒ θ = $\frac{3\pi }{4}$

⇒  $sin$θ $=sin \frac{3\pi }{4}$

            $=\frac{1}{\sqrt{2} }$

⇒ $sec$ θ $=sec \frac{3\pi }{4}$

            $=-\sqrt{2}$

Answer 2

Given : sinθsecθ= -1   and θ lies in the second quadrant

To Find : sinθ and  secθ

Solution:

sinθsecθ= -1

=> sinθ/cosθ= -1

=> tanθ= -1

=> θ = 135° as  θ lies in the second quadrant

Sin 135°  = 1/√2

cos 135°  = -1/√2

=> sec  135°  = - √2

sinθ  = 1/√2 and  secθ =  - √2

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