If sec θ = √2 and \(\frac{3π}{2}\) < θ < 2π , find the value of�

1.	If sec θ = √2 and \(\frac{3π}{2}\) < θ < 2π , find the value of \(\frac{\text{1 tan θ + cosec θ }}{\text{1 cot θ – cosec θ}}.\)
Answer» sec θ = √2 ⇒ cos θ = \(\frac1{\sqrt2}\) ∴ sin θ = ±\(\sqrt{1-cos^2\theta}\) = ±\(\sqrt{1-\frac12}\) = ±\(\frac1{\sqrt2}\) Since θ lies in the fourth quadrant, so sin θ is –ve and cos q is +ve. ∴ sin θ = \(-\frac1{\sqrt2}\), cosec θ = √2 tan θ = \(\frac{sin\,\theta}{cos\,\theta}\) = \(-\frac1{\sqrt2}\) x \(\frac{\sqrt2}{1}\) = –1 ⇒ cot θ = –1 ∴ \(\frac{1+tan\,\theta+cosec\,\theta}{1+cot\,\theta-cosec\,\theta}\) = \(\frac{1-1-\sqrt2}{1-1+\sqrt2}= -1.\)

If sec θ = √2 and \(\frac{3π}{2}\) < θ < 2π , find the value of \(\frac{\text{1 tan θ + cosec θ }}{\text{1 cot θ – cosec θ}}.\)

Answer»

sec θ = √2 ⇒ cos θ = \(\frac1{\sqrt2}\)

∴ sin θ = ±\(\sqrt{1-cos^2\theta}\) = ±\(\sqrt{1-\frac12}\) = ±\(\frac1{\sqrt2}\)

Since θ lies in the fourth quadrant, so sin θ is –ve and cos q is +ve.

∴ sin θ = \(-\frac1{\sqrt2}\), cosec θ = √2

tan θ = \(\frac{sin\,\theta}{cos\,\theta}\) = \(-\frac1{\sqrt2}\) x \(\frac{\sqrt2}{1}\) = –1 ⇒ cot θ = –1

∴ \(\frac{1+tan\,\theta+cosec\,\theta}{1+cot\,\theta-cosec\,\theta}\) = \(\frac{1-1-\sqrt2}{1-1+\sqrt2}= -1.\)