Find the value of \(\rm \sqrt{\frac{1-\sin 3\theta}{1+\sin 3\theta}} \)

Calculation:

\(\rm \sqrt{\frac{1-\sin 3\theta}{1+\sin 3\theta}} \)

⇒ By putting θ = 15° 

\(\rm \sqrt{\frac{1-\sin 3\theta}{1+\sin 3\theta}} \)

⇒ \(\rm \sqrt{\frac{1-\sin 45}{1+\sin 45}} \)

⇒ \(\rm \sqrt{\frac{1-\ (1/\sqrt2)}{1+\ (1/\sqrt2)}} \) = \(\rm \sqrt{\frac{\sqrt2 \ - \ 1}{\sqrt2\ + \ 1}} \)

⇒ \(\rm \sqrt{\frac{\sqrt2 \ - \ 1}{\sqrt2\ + \ 1} \times\frac{\sqrt2 \ - \ 1}{\sqrt2\ - \ 1}} \)

⇒ \(\rm \sqrt{\frac{(\sqrt2 \ - \ 1)^2}{2-1}} \)

⇒ \(\rm \sqrt{(\sqrt2 \ - \ 1)^2}\)

⇒ √2 - 1

Again, In option 1  By putting θ = 15° we get,

sec 3θ - tan 3θ = sec 45° - tan 45°

⇒ √2 - 1

∴ The correct option is 1

Alternate Method

\(\rm \sqrt{\frac{1-\sin 3\theta}{1+\sin 3\theta}} \)

⇒ \(\rm \sqrt{{\frac{1-\sin 3\theta}{1+\sin 3\theta}} \times\frac{1-sin3\theta}{1-sin3\theta}} \)

⇒ \(\rm \sqrt{\frac{(1-\sin 3\theta)^2}{1^2-\sin^2 3\theta}} \)

⇒ \(\rm \sqrt{\frac{(1-\sin 3\theta)^2}{cos^2 3\theta}} \)

⇒ \(\rm \sqrt{(\frac{1-\sin 3\theta}{cos 3\theta}})^2 \)

⇒ \(\frac{1 - sin3\theta}{cos3\theta}\)

⇒ \(\frac{1}{cos3\theta}-\frac{sin3\theta}{cos3\theta}\)

⇒ sec 3θ - tan 3θ

∴ The correct option is 1