Question

Please help me with this question.​

Answer 1

Step-by-step explanation:

here the answer any doubt ask me

Answer 2

Answer 1:

${: \implies \dfrac{ \cos \theta}{1 + \sin \theta } + \dfrac{ \cos \theta }{1 - \sin \theta } = 2 \sec \theta}$

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» Taking LCM

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${:\implies \dfrac{ \cos \theta(1 - \sin \theta) + \cos \theta(1 - \sin \theta) }{(1 + \sin \theta )(1 - \sin \theta)} = 2 \sec \theta}$

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» Apply identity (A+B)(A-B)=A²-B²

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${:\implies \dfrac{ \cos \theta(1 - \sin \theta) + \cos \theta(1 + \sin \theta) }{ (1 - \sin ^{2} \theta)} = 2 \sec \theta}$

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» Apply identity 1 - sin² θ= cos² θ

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${:\implies \dfrac{ \cos \theta(1 - \sin \theta) + \cos \theta(1 + \sin \theta) }{ ( \cos ^{2} \theta)} = 2 \sec \theta}$

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» Taking cos θ common in nr.

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${:\implies \dfrac{ \cos \theta(1 - \sin \theta + 1 - \sin \theta) }{ ( \cos ^{2} \theta)} = 2 \sec \theta}$

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${:\implies \dfrac{ \cos \theta(2) }{ ( \cos ^{2} \theta)} = 2 \sec \theta}$

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${:\implies \dfrac{ \cancel{ \cos \theta}(2) }{ ( \cancel{ \cos ^{2} \theta)}} = 2 \sec \theta}$

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${:\implies \dfrac{ 2}{ ( \cos \theta)} = 2 \sec \theta}$

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» Apply formula 1/cos θ = sec θ

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${:\implies 2 \sec \theta = 2 \sec \theta}$

Hence proved

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$\rule{240}{1}$

Answer 2:

${:\implies \dfrac{1 + \cos \theta}{ \sin \theta} + \dfrac{ \sin \theta}{1 + \cos \theta } = 2 \: { \text{cosec}} \: \theta}$

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» Taking LCM

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${:\implies \dfrac{( 1 + \cos \theta)^{2} + \sin ^{2} \theta}{(\sin \theta)(1 + \cos \theta) } = 2 \: { \text{cosec}} \: \theta}$

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» Apply identity (A+B)²=A²+B²+2AB

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${:\implies \dfrac{ 1 +2 \cos \theta + \cos ^{2} \theta + \sin ^{2} \theta}{(\sin \theta)(1 + \cos \theta) } = 2 \: { \text{cosec}} \: \theta}$

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» Apply formula sin² θ + cos² θ = 1

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${:\implies \dfrac{ 1 +2 \cos \theta + 1}{(\sin \theta)(1 + \cos \theta) } = 2 \: { \text{cosec}} \: \theta}$

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${:\implies \dfrac{ 2 +2 \cos \theta }{(\sin \theta)(1 + \cos \theta) } = 2 \: { \text{cosec}} \: \theta}$

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${:\implies \dfrac{ 2( 1+ \cos \theta) }{(\sin \theta)(1 + \cos \theta) } = 2 \: { \text{cosec}} \: \theta}$

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${:\implies \dfrac{ 2 \cancel{( 1+ \cos \theta)} }{(\sin \theta) \cancel{(1 + \cos \theta)} } = 2 \: { \text{cosec}} \: \theta}$

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${:\implies \dfrac{ 2 }{\sin \theta } = 2 \: { \text{cosec}} \: \theta}$

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» Apply formula 1/sin θ= cosec θ

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${:\implies 2 \: { \text{cosec}} \: \theta = 2 \: { \text{cosec}} \: \theta}$

Hence proved