Question

Prove that 1+cosA+sinA/1+cosA-sinA=1+sinA/cosA

Answer 1

Step-by-step explanation:

Refer to the attachment.

Some identities :

• 1/SecA = CosA

• 1/CosA = SecA

• SinA/CosA = tanA

• cosA/sinA = cotA

• 1/sinA = CosecA

• CosecA = 1/SinA

• Sec²A - Tan²A = 1

• Cosec²A - Cot²A = 1

• Sin²A + Cos²A = 1

• (a+b)(a-b) = a² - b²

Answer 2

To Prove

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

Proof

LHS

$\sf \dfrac{1 + \cos(x) + \sin(x) }{1 + \cos(x) - \sin(x) } \:$

Dividing both numerator and denominator by cos(x)

$\longrightarrow \: \sf \: \dfrac{ \sec(x) + \tan(x) + 1 }{ \sec(x) - \tan(x) + 1 }$

But sec²∅ - tan²∅ = 1

$\longrightarrow \: \tt \: \: \dfrac{ \sec(x) + \tan(x) + \bigg({sec}^{2} (x) - {tan}^{2} (x) \bigg)}{\sec(x) + \tan(x) - 1}$

Now,

sec²x - tan²x = (sec x + tan x)(sec x - tan x )

Thus,

$\longrightarrow \: \sf \: \dfrac{ \sec(x) - \tan(x) + 1 }{ \sec(x) - \tan(x) + 1 } \times \sec(x ) + \tan(x) \\ \\ \longrightarrow \: \sf \: \dfrac{1}{ \cos(x) } + \dfrac{ \sin(x) }{ \cos(x) } \\ \\ \longrightarrow \: \sf \: \dfrac{1 + \sin( x) }{ \cos(x) }$