Question

√[(1+sinA)/(1-sinA)]=?​

Answer 1

Answer:

answer for the given problem is given

Answer 2

Answer:

secA + tanA

Explanation:

$\sf \Longrightarrow \sqrt{\dfrac{1+sinA}{1-sinA}}$

$\sf \Longrightarrow \sqrt{\dfrac{1+sinA}{1-sinA} \times \dfrac{1+sinA}{1+sinA}}$

Identities used:

(a + b)(a + b) = (a + b)

(a + b)² = a² + b² + 2ab

(a + b)(a - b) = a² - b²

$\sf \Longrightarrow \sqrt{\dfrac{\Big(1+sinA\Big)^2}{\Big(1\Big)^2-\Big(sinA\Big)^2}}$

$\sf \Longrightarrow \sqrt{\dfrac{ \ \Big(1+sinA\Big)^2}{1-sin^2A}}$

Trigonometric Identity used:

1 - sin²A = cos²A

$\sf \Longrightarrow \sqrt{\dfrac{ \ \Big(1+sinA\Big)^2}{cos^2A}}$

Squares and roots get cancelled.

$\sf \Longrightarrow \dfrac{1+sinA}{cosA}$

$\sf \Longrightarrow \dfrac{1}{cosA} + \dfrac{sinA}{cosA}$

Trigonometric ratios used:

1/cosA = secA

sinA/cosA = tanA

$\sf \Longrightarrow secA + tanA$

Final answer: secA + tanA