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

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

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Mathematics Ayla 11 months 2021-06-30T16:07:08+00:00 2021-06-30T16:07:08+00:00 2 Answers 0 views 0

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Captcha* 9:3-3+1x3-4:2 = ? ( )

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Lyla · Answer 1 · 2021-06-30T16:08:49+00:00

Lyla

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2021-06-30T16:08:49+00:00 June 30, 2021 at 4:08 pm

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Missbelladonna answer is correct

Luna · Answer 2 · 2021-06-30T16:09:07+00:00

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 ]

⠀⠀⠀⠀⠀━━━━━━━━━━━━━━━━━━━⠀

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