Question

If sin theta + cos theta = m and sec theta + cosec theta = n , prove that n(m² – 1) = 2m.

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Answer 1

✬ LHS = RHS ✬

Step-by-step explanation:

Given:

• sinθ + cosθ is equal to m.
• secθ + cosecθ is equal to n.

To Prove:

• n(m² – 1) = 2m

Proof: Putting all the values in LHS.

➟ n(m² – 1)

➟ (secθ + cosecθ) [(sinθ + cosθ)² – 1]

➟ (1/cosθ + 1/sinθ) [(sin²θ + cos²θ + 2sinθcosθ) – 1]

➟ (sinθ + cosθ/sinθcosθ) [1 + 2sinθcosθ – 1]

➟ (sinθ + cosθ/sinθcosθ) × 2sinθcosθ

➟ 2 × sinθ + cosθ

➟ 2 × m

➟ 2m

Hence, LHS = RHS

★ Formulae used

• secθ = 1/cosθ

• sinθ = 1/cosecθ

• cosecθ = 1/sinθ

• sin²θ + cos²θ = 1

Answer 2

Given : sin θ + cos θ = m , and sec θ + cosec θ= n .

Exigency To Prove : n(m² – 1) = 2m . [ L.H.S = R.H.S ]

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⠀⠀⠀Given that ,

⠀⠀⠀⠀⠀⠀▪︎⠀$\sf \: sin \:\theta \: + \: \: cos \:\theta \: =\: m \:$

⠀⠀⠀⠀⠀⠀▪︎⠀$\sf \: sec \:\theta \: + \: \: cosec \:\theta \: =\: n \:$

⠀⠀⠀Need To Proove :

⠀⠀⠀⠀⠀⠀▪︎⠀$\bf{ \: n \:\bigg( \: m^2\: - \:1 \:\bigg) \: =\: 2m \: } \:$

⠀⠀Here ,

⠀⠀⠀⠀⠀⠀$\leadsto \: \bf L.H.S \:\:\sf =\: \: n \:\bigg( \: m^2\: - \:1 \:\bigg) \:$

⠀⠀⠀⠀⠀⠀$\leadsto \: \bf R.H.S \:\:\sf =\: \: 2m \:$

$\qquad \bigstar \underline {\boldsymbol {Now \: by \:taking \: the \: \:L.H.S \:\: :}}$

⠀⠀⠀⠀⠀⠀$\leadsto \: \bf L.H.S \:\:\sf =\: \: n \:\bigg( \: m^2\: - \:1 \:\bigg) \:$

$\qquad \dashrightarrow \:\sf \: \: n \:\bigg( \: m^2\: - \:1 \:\bigg) \:$

⠀⠀⠀⠀⠀⠀$\underline {\boldsymbol{\star\:Now \: By \: Substituting \: the \: given \: Values \::}}$

$\qquad \dashrightarrow \:\sf \: \: n \:\bigg( \: m^2\: - \:1 \:\bigg) \:$

⠀⠀⠀▪︎⠀$\sf \: sin \:\theta \: + \: \: cos \:\theta \: =\: m \:$

⠀⠀⠀▪︎⠀$\sf \: sec \:\theta \: + \: \: cosec \:\theta \: =\: n \:$

$\qquad \dashrightarrow \:\sf \: \: n \:\bigg( \: m^2\: - \:1 \:\bigg) \:$

$\qquad \dashrightarrow \:\sf \: \bigg(\:sec \:\theta \: + \: \: cosec \:\theta \: \bigg) \: \:\bigg[ \: \bigg(\:sin \:\theta \: + \: \: cos \:\theta \: \bigg)^2\: - \:1 \:\bigg] \:$

$\qquad \because\:\:\bigg\lgroup \sf{ \red{ sec \: \theta \: =\:\dfrac{1}{\:cos \:\theta }}}\bigg\rgroup$

⠀⠀⠀⠀⠀AND ,

$\qquad \because\:\:\bigg\lgroup \sf{ \red{ cosec \: \theta \: =\:\dfrac{1}{\:sin \:\theta }}}\bigg\rgroup$

$\qquad \dashrightarrow \:\sf \: \bigg(\:sec \:\theta \: + \: \: cosec \:\theta \: \bigg) \: \:\bigg[ \: \bigg(\:sin \:\theta \: + \: \: cos \:\theta \: \bigg)^2\: - \:1 \:\bigg] \:$

$\qquad \dashrightarrow \:\sf \: \bigg(\:\dfrac{1}{cos \:\theta } \: + \: \: \dfrac{1}{cos \:\theta} \: \bigg) \: \:\bigg[ \: \bigg(\:sin \:\theta \: + \: \: cos \:\theta \: \bigg)^2\: - \:1 \:\bigg] \:$

$\qquad \dashrightarrow \:\sf \: \bigg(\:\dfrac{\: sin \:\theta \: + \: \: cos \:\theta \: }{cos \:\theta\:\sin \:\theta } \: \: \bigg) \: \:\bigg[ \: \bigg(\:sin \:\theta \: + \: \: cos \:\theta \: \bigg)^2\: - \:1 \:\bigg] \:$

$\qquad \because\:\:\bigg\lgroup \sf{ \red{ ( \: a +\:b)^2\:\:=\:a^2 + b^2 + 2ab }}\bigg\rgroup$

$\qquad \dashrightarrow \:\sf \: \bigg(\:\dfrac{\: sin \:\theta \: + \: \: cos \:\theta \: }{cos \:\theta\:\sin \:\theta } \: \: \bigg) \: \:\bigg[ \: \:sin^2 \:\theta \: + \: \: cos^2 \:\theta \:+\: 2cos \:\theta\:\sin \:\theta \: - \:1 \:\bigg] \:$

$\qquad \because\:\:\bigg\lgroup \sf{ \red{ sin^2 \: \theta \: +\:\:cos^2 \:\theta\:\:=\:1 }}\bigg\rgroup$

$\qquad \dashrightarrow \:\sf \: \bigg(\:\dfrac{\: sin \:\theta \: + \: \: cos \:\theta \: }{cos \:\theta\:\sin \:\theta } \: \: \bigg) \: \:\bigg[ \: \:sin^2 \:\theta \: + \: \: cos^2 \:\theta \:+\: 2cos \:\theta\:\sin \:\theta \: - \:1 \:\bigg] \:$

$\qquad \dashrightarrow \:\sf \: \bigg(\:\dfrac{\: sin \:\theta \: + \: \: cos \:\theta \: }{cos \:\theta\:\sin \:\theta } \: \: \bigg) \: \:\bigg[ \: \:1 \:+\: 2cos \:\theta\:\sin \:\theta \: - \:1 \:\bigg] \:$

$\qquad \dashrightarrow \:\sf \: \bigg(\:\dfrac{\: sin \:\theta \: + \: \: cos \:\theta \: }{cos \:\theta\:\sin \:\theta } \: \: \bigg) \: \:\bigg[ \: \cancel {\:1 }\:+\: 2cos \:\theta\:\sin \:\theta \:\cancel {- \:1 }\:\bigg] \:$

$\qquad \dashrightarrow \:\sf \: \bigg(\:\dfrac{\: sin \:\theta \: + \: \: cos \:\theta \: }{cos \:\theta\:\sin \:\theta } \: \: \bigg) \: \:\bigg[ \: \:\: 2\:cos \:\theta\:\sin \:\theta \: \:\bigg] \:$

$\qquad \dashrightarrow \:\sf \: \bigg(\:\dfrac{\: sin \:\theta \: + \: \: cos \:\theta \: }{\cancel{ cos \:\theta\:\sin \:\theta} } \: \: \bigg) \: \:\bigg[ \: \:\: 2\:\cancel {\:\:cos \:\theta\:\sin \:\theta \: } \:\bigg] \:\\\\\qquad \dashrightarrow \:\sf \: \bigg( sin \:\theta \: + \: \: cos \:\theta \: \bigg) \: \:\bigg( \: \:\: 2\: \: \:\bigg) \:\\\\\qquad \dashrightarrow \:\sf \:2 \times \bigg( sin \:\theta \: + \: \: cos \:\theta \: \bigg) \: \: \:$

⠀⠀⠀▪︎⠀$\sf \: sin \:\theta \: + \: \: cos \:\theta \: =\: m \:$

$\qquad \dashrightarrow \:\sf \:2 \times \bigg( sin \:\theta \: + \: \: cos \:\theta \: \bigg) \: \: \:\\\\\qquad \dashrightarrow \:\sf \:2 \times \bigg( m\: \bigg) \: \: \:\\\\\qquad \dashrightarrow \:\sf \:2 \times m\: \: \:\\\\\qquad \dashrightarrow \:\sf \:2m\: \: \:\\\\\qquad \therefore \:\:\pmb{\underline{\purple{\frak{\:L.H.S \:\:=\:\: 2m \: }}} }\:\:\bigstar$

⠀⠀⠀⠀⠀$\leadsto \: \bf L.H.S \:\:\sf =\: \: 2m \:$

⠀⠀⠀⠀⠀$\leadsto \: \bf R.H.S \:\:\sf =\: \: 2m \:$

⠀⠀As , We can see that here , [ L.H.S = R.H.S ]

$\therefore {\underline {\bf{ Hence, \:Proved\:}}}$

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