Question

Cos A/ 1+sin A+ 1+sinA/ cosA= 2sec A

Answer 1

Solution:

We have to prove that:

$\rm \longrightarrow \dfrac{ \cos(x) }{1 + \sin(x) } +\dfrac{1 + \sin(x) }{ \cos(x) } = 2 \sec(x)$

Taking Left Hand Side, we get:

$\rm = \dfrac{ \cos(x) }{1 + \sin(x) } +\dfrac{1 + \sin(x) }{ \cos(x) }$

$\rm = \dfrac{ \cos^{2} (x) + { \{ 1 + \sin(x) \}}^{2} }{ \{1 + \sin(x) \} \{ \cos(x) \} }$

$\rm = \dfrac{ \cos^{2} (x) +1 + 2 \sin(x) + \sin^{2}(x)}{ \{1 + \sin(x) \} \{ \cos(x) \} }$

$\rm = \dfrac{ \cos^{2} (x) + \sin^{2} (x) +1 + 2 \sin(x)}{ \{1 + \sin(x) \} \{ \cos(x) \} }$

We know that:

$\rm \longrightarrow \cos^{2} (x) + \sin^{2} (x) =1$

So, LHS becomes:

$\rm = \dfrac{1+1 + 2 \sin(x)}{ \{1 + \sin(x) \} \{ \cos(x) \} }$

$\rm = \dfrac{2+ 2 \sin(x)}{ \{1 + \sin(x) \} \{ \cos(x) \} }$

$\rm = \dfrac{2 \{1+ \sin(x) \}}{ \{1 + \sin(x) \} \{ \cos(x) \} }$

$\rm = \dfrac{2}{\cos(x)}$

$\rm = 2 \sec(x)$

Therefore:

$\rm \longrightarrow \dfrac{ \cos(x) }{1 + \sin(x) } +\dfrac{1 + \sin(x) }{ \cos(x) } = 2 \sec(x)$

Hence Proved..!!

Additional Information:

1. Relationship between sides and T-Ratios.

• sin θ = Height/Hypotenuse
• cos θ = Base/Hypotenuse
• tan θ = Height/Base
• cot θ = Base/Height
• sec θ = Hypotenuse/Base
• cosec θ = Hypotenuse/Height

2. Square formulae.

• sin²θ + cos²θ = 1
• cosec²θ - cot²θ = 1
• sec²θ - tan²θ = 1

3. Reciprocal Relationship.

• sin θ = 1/cosec θ
• cos θ = 1/sec θ
• tan θ = 1/cot θ
• cosec θ = 1/sin θ
• sec θ = 1/cos θ
• tan θ = 1/cot θ

4. Cofunction identities.

• sin(90° - θ) = cos θ
• cos(90° - θ) = sin θ
• cosec(90° - θ) = sec θ
• sec(90° - θ) = cosec θ
• tan(90° - θ) = cot θ
• cot(90° - θ) = tan θ

5. Even odd identities.

• sin -θ = -sin θ
• cos -θ = cos θ
• tan -θ = -tan θ