If secθ+cosec θ=2√2 , where 0^°<θ<90^° , then what is the value of θ ?

The correct answer is Option (1) → \(45^\circ\)

At \(45^\circ\) the sine and cosine are equal, so \(\sec 45^\circ=\mathrm{cosec}\,45^\circ=\sqrt{2}\), and their sum is \(2\sqrt{2}\).

• \(\theta=45^\circ\): \(\sqrt{2}+\sqrt{2}=2\sqrt{2}\)

• No other acute angle works: \(\sin\theta\cos\theta=\frac{1}{2}\sin 2\theta\le\frac{1}{2}\), so \(\sec\theta\,\mathrm{cosec}\,\theta\ge 2\)

• The sum of two positive numbers is at least twice the square root of their product: \(\sec\theta+\mathrm{cosec}\,\theta\ge 2\sqrt{2}\), with equality only when \(\sec\theta=\mathrm{cosec}\,\theta\), that is at \(\theta=45^\circ\)

Why the other options are incorrect

• Option (2) \(15^\circ\) — Incorrect. \(\sec 15^\circ+\mathrm{cosec}\,15^\circ\approx 1.04+3.86=4.90\), well above \(2\sqrt{2}\approx 2.83\).

• Option (3) \(30^\circ\) — Incorrect. \(\sec 30^\circ+\mathrm{cosec}\,30^\circ=\frac{2}{\sqrt{3}}+2\approx 3.15\).

• Option (4) \(60^\circ\) — Incorrect. \(\sec 60^\circ+\mathrm{cosec}\,60^\circ=2+\frac{2}{\sqrt{3}}\approx 3.15\), the same sum as at \(30^\circ\).