Question

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Answer 1

$\sf{Given : \dfrac{1 + CosA}{SinA}}$

$\textsf{Multiplying the Numerator and Denominator with (1 - CosA), We get :}$

$\implies \sf{\dfrac{(1 + CosA)(1 - CosA)}{SinA(1 - CosA)}$

$\textsf{We know that : \;} \boxed{\sf{(a + b)(a - b) = a^2 - b^2}}$

$\implies \boxed{\sf{(1 + CosA)(1 - CosA) = 1 - Cos^2A}}$

$\implies \sf{\dfrac{(1 - Cos^2A)}{SinA(1 - CosA)}$

$\textsf{We know that : \;} \boxed{\sf{1 - Cos^2A = Sin^2A}}$

$\implies \sf{\dfrac{Sin^2A}{SinA(1 - CosA)}$

$\implies \sf{\dfrac{SinA}{(1 - CosA)}$

Answer 2

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$\underline{\huge{\textsf{To Prove : }}}$

$\dfrac{1 + cos\, A}{sin\, A} = \dfrac{sin\, A}{1-cos\, A}$

$\underline{\underline{\huge{\textbf{Solution:}}}}$

$\underline{\large{\textbf{Step-1}}}$
Consider LHS

$\underline\large\textbf{LHS =}}}$

$= \frac{1 + cos\, A}{sin\, A}$

$\underline{\large{\textbf{Step-2}}}$
Rationalize the Numerator.

i.e., Multiplying and Dividing with $(1 + cos\, A)$

$\implies \frac{(1 + cos\, A)(1-cos\, A)}{sin\, A(1-cos\, A)}$

$\underline{\large{\textbf{Step-3}}}$
Express numerator, Using the identity
$(a+b)(a-b)= (a^{2}-b^{2})$

$(1 + cos\, A)(1-cos\, A) = (1-cos^{2}\, A)$

Substitute in Numerator.

$\implies \frac{1 -cos^{2}\, A}{sin\, A(1-cos\, A)}$

$\underline{\large{\textbf{Step-4}}}$
Express the Numerator, Using the Identity
$sin^{2}\, A =(1-cos^{2}\, A)$

Substitute in Numerator

$\implies \frac{sin^{2}\, A}{sin\, A(1-cos\, A)}$

$\underline{\large{\textbf{Step-5}}}$
cancel out the common sin A in Numerator and Denominator.

$\implies \frac{sin\, A}{1-cos\, A}$

$\underline\large\textbf{=RHS}}}$

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