\(\frac{{\cos x}}{{1 + \sin x}} + \frac{{1 + \sin x}}{{\cos x}}\;\) i

1.	\(\frac{{\cos x}}{{1 + \sin x}} + \frac{{1 + \sin x}}{{\cos x}}\;\) is equal to:1. 2cos x2. 2sin x3. 2sec x4. 2cosec x
Answer» Correct Answer - Option 3 : 2sec x Given: \( \frac{{\cos x}}{{1 + \sin x}} + \frac{{1 + \sin x}}{{\cos x}}\) Concept Used: sin2x + cos2x = 1 Calculation: \( \frac{{\cos x}}{{1 + \sin x}} + \frac{{1 + \sin x}}{{\cos x}}\) ⇒ {cos²x + (1 + sinx)²}/{(1 + sinx) × cosx} ⇒ (cos²x + 1 + sin²x + 2sinx)/{(1 + sinx) × cosx} ⇒ (1 + 1 + 2sinx)/{(1 + sinx) × cosx} [∵ sin²x + cos²x = 1] ⇒ 2(1 + sinx)/{(1 + sinx) × cosx} ⇒ 2/cosx ⇒ 2secx ∴ \( \frac{{\cos x}}{{1 + \sin x}} + \frac{{1 + \sin x}}{{\cos x}}\) = 2secx

\(\frac{{\cos x}}{{1 + \sin x}} + \frac{{1 + \sin x}}{{\cos x}}\;\) is equal to:1. 2cos x2. 2sin x3. 2sec x4. 2cosec x

Answer» Correct Answer - Option 3 : 2sec x

Given:

\( \frac{{\cos x}}{{1 + \sin x}} + \frac{{1 + \sin x}}{{\cos x}}\)

Concept Used:

sin2x + cos2x = 1

Calculation:

\( \frac{{\cos x}}{{1 + \sin x}} + \frac{{1 + \sin x}}{{\cos x}}\)

⇒ {cos²x + (1 + sinx)²}/{(1 + sinx) × cosx}

⇒ (cos²x + 1 + sin²x + 2sinx)/{(1 + sinx) × cosx}

⇒ (1 + 1 + 2sinx)/{(1 + sinx) × cosx} [∵ sin²x + cos²x = 1]

⇒ 2(1 + sinx)/{(1 + sinx) × cosx}

⇒ 2/cosx

⇒ 2secx

∴ \( \frac{{\cos x}}{{1 + \sin x}} + \frac{{1 + \sin x}}{{\cos x}}\) = 2secx