Question

Sin A/ 1+Cos A + 1 + CosA/ sinA​

Answer 1

To Find:

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

Taking LCM we get:

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

Using (a + b)(a + b) = (a + b)² we get:

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

Using (a + b)² = a² + b² + 2ab we get:

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

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

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

Using the identity sin²A + cos²A = 1 we get:

$\Longrightarrow \sf \dfrac{1 + 1 + 2cosA}{(1 + cosA)(sinA)}$

$\Longrightarrow \sf \dfrac{2 + 2cosA}{(1 + cosA)(sinA)}$

Take 2 out since it's common to both 2 and 2cosA.

$\Longrightarrow \sf \dfrac{2(1 + cosA)}{(1 + cosA)(sinA)}$

Cancelling (1 + cosA) in the numerator & denominator we get:

$\Longrightarrow \sf \dfrac{2}{sinA}$

$\Longrightarrow \sf 2 \times \dfrac{1}{sinA}$

Using 1/sinA = cosecA

$\Longrightarrow \sf 2cosecA$

Hence solved.

Answer 2

Taking LCM we get:

\Longrightarrow \sf \dfrac{(sinA)(sinA) + (1 + cosA)(1 + cosA)}{(1 + cosA)(sinA)}⟹

(1+cosA)(sinA)

(sinA)(sinA)+(1+cosA)(1+cosA)

Using (a + b)(a + b) = (a + b)² we get:

\Longrightarrow \sf \dfrac{sin^2A + (1 + cosA)^2}{(1 + cosA)(sinA)}⟹

(1+cosA)(sinA)

sin

2

A+(1+cosA)

2

Using (a + b)² = a² + b² + 2ab we get:

\Longrightarrow \sf \dfrac{sin^2A + (1)^{2} + (cosA)^2 + 2(1)(cosA)}{(1 + cosA)(sinA)}⟹

(1+cosA)(sinA)

sin

2

A+(1)

2

+(cosA)

2

+2(1)(cosA)

\Longrightarrow \sf \dfrac{sin^2A + 1 + cos^2A + 2cosA}{(1 + cosA)(sinA)}⟹

(1+cosA)(sinA)

sin

2

A+1+cos

2

A+2cosA

\Longrightarrow \sf \dfrac{sin^2A + cos^2A + 1 + 2cosA}{(1 + cosA)(sinA)}⟹

(1+cosA)(sinA)

sin

2

A+cos

2

A+1+2cosA

Using the identity sin²A + cos²A = 1 we get:

\Longrightarrow \sf \dfrac{1 + 1 + 2cosA}{(1 + cosA)(sinA)}⟹

(1+cosA)(sinA)

1+1+2cosA

\Longrightarrow \sf \dfrac{2 + 2cosA}{(1 + cosA)(sinA)}⟹

(1+cosA)(sinA)

2+2cosA

Take 2 out since it's common to both 2 and 2cosA.

\Longrightarrow \sf \dfrac{2(1 + cosA)}{(1 + cosA)(sinA)}⟹

(1+cosA)(sinA)

2(1+cosA)

Cancelling (1 + cosA) in the numerator & denominator we get:

\Longrightarrow \sf \dfrac{2}{sinA}⟹

sinA

2

\Longrightarrow \sf 2 \times \dfrac{1}{sinA}⟹2×

sinA

1

Using 1/sinA = cosecA

\Longrightarrow \sf 2cosecA⟹2cosecA

Hence solved.