Given \(0^\circ<\theta<90^\circ\), the positive square root \(\sqrt{\displaystyle\frac{1+\sin\theta}{1-\sin\theta}}\) can be written more simply as:
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → \(\sec\theta+\tan\theta\)
Multiply the top and the bottom inside the root by \(1+\sin\theta\). The bottom becomes \(1-\sin^2\theta=\cos^2\theta\), a perfect square, so the root can be taken. For \(0^\circ<\theta<90^\circ\), \(\cos\theta>0\) and \(1+\sin\theta>0\), so the root of each square is the quantity itself.
\(\displaystyle\frac{1+\sin\theta}{1-\sin\theta}=\frac{(1+\sin\theta)^2}{1-\sin^2\theta}=\frac{(1+\sin\theta)^2}{\cos^2\theta}\)
Square root \(=\displaystyle\frac{1+\sin\theta}{\cos\theta}\)
\(\displaystyle\frac{1}{\cos\theta}+\frac{\sin\theta}{\cos\theta}=\sec\theta+\tan\theta\)