Question

prove that question​

Answer 1

Answer:

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Answer 2

Question :-

Prove that

$\frac{ \sin \theta}{1 + \cos \theta} + \frac{1 + \cos \theta}{ \sin \theta} = 2 \csc \theta$

Proof :-

→ We will have to show that left hand side equal to right hand side .

Formula used :-

$\boxed{ \green{\star }} \: \: { \sin}^{2} \theta \: + { \cos}^{2} \theta \: = 1 \\ \\ \boxed{ \red{\star }}\: \frac{1}{ \sin \theta} = \csc \theta \: \\ \\ \boxed{ \orange{\star }}\: \sf{ {x}^{2} + {y}^{2} + 2xy = {(x + y)}^{2} }$

Explanation :-

Firstly take left hand side (LHS)

$\rightarrow \: \frac{ \sin \theta}{1 + \cos \theta} + \frac{1 + \cos \theta}{ \sin \theta} \: \\ \\ \rightarrow \: \frac{ { \sin}^{2} \theta + {(1 + \cos \theta)}^{2} }{ \sin \theta \: (1 + \cos \theta)} \\ \\ \because \sf{ {(x + y)}^{2} = {x}^{2} + {y}^{2} + 2xy} \\ \\ \rightarrow \: \frac{ { \sin}^{2} \theta \: + {(1)}^{2} + { \cos}^{2} \theta + 2 \cos \theta }{ \sin \theta(1 + \cos \theta)} \\ \because \: { \sin}^{2} \theta \: + { \cos}^{2} \theta \: = 1 \\ \therefore \: \\ \\ \rightarrow \: \frac{1 + 1 + 2 \cos \theta}{ \sin \theta \: (1 + \cos \theta \: )} \\ \\ \rightarrow \: \frac{2 + 2 \cos \theta}{ \sin \theta \: (1 + \cos \theta)} \\ \\ \rightarrow \: \frac{2(1 + \cos \theta)}{ \sin \theta \: (1 + \cos \theta)} \\ \\ \rightarrow \: 2 \times \frac{1}{ \sin \theta} \\ \\ \because \: \csc \theta \: = \frac{1}{ \sin \theta} \\ \\ \rightarrow \: 2 \: \csc \theta \:$

Left hand side = right hand side

(LHS = RHS )

hence proved : )

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