Question

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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

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

Take LCM to denominator

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

Using ( a +b)² formula for (1+cosA)²

(a + b)² = a² + 2ab + b²

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

As, we know that sin²A + cos²A = 1

For denominator multiply with sinA

$\sf\dfrac{sin^2A + cos^2A + 1 +2cosA}{sinA+sinAcosA}$

$\sf\dfrac{1+1+2cosA}{sinA + sinAcosA}$

$\sf\dfrac{2+2cosA}{sinA+sinAcosA}$

Take common to numerator 2 & take common denominator sinA

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

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

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

We know that,

1/sinA = cosecA

2 $\sf\dfrac{1}{sinA}$ = 2cosecA

So,

$\sf\dfrac{sinA}{1+cosA}$ + $\sf\dfrac{1+cosA}{sinA}$ = 2cosecA

Used formulae:-

sin² A + cos²A =1

(a + b )² = a² + 2ab + b²

1/sinA = cosecA

Know more :-

Trignometric Identities

sin²θ + cos²θ = 1

sec²θ - tan²θ = 1

csc²θ - cot²θ = 1

Trignometric relations

sinθ = 1/cscθ

cosθ = 1 /secθ

tanθ = 1/cotθ

tanθ = sinθ/cosθ

cotθ = cosθ/sinθ

Trignometric ratios

sinθ = opp/hyp

cosθ = adj/hyp

tanθ = opp/adj

cotθ = adj/opp

cscθ = hyp/opp

secθ = hyp/adj