Question

can anyone solve this plz​

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

Answer:

[]To Find:

\Longrightarrow \sf \dfrac{sinA}{1 + cosA} + \dfrac{1 + cosA}{sinA}⟹1+cosAsinA+sinA1+cosA

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)sin2A+(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)sin2A+(1)2+(cosA)2+2(1)(cosA)

\Longrightarrow \sf \dfrac{sin^2A + 1 + cos^2A + 2cosA}{(1 + cosA)(sinA)}⟹(1+cosA)(sinA)sin2A+1+cos2A+2cosA

\Longrightarrow \sf \dfrac{sin^2A + cos^2A + 1 + 2cosA}{(1 + cosA)(sinA)}⟹(1+cosA)(sinA)sin2A+cos2A+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}⟹sinA2

\Longrightarrow \sf 2 \times \dfrac{1}{sinA}⟹2×sinA1

Using 1/sinA = cosecA

\Longrightarrow \sf 2cosecA⟹2cosecA

Hence solved.

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