If sin theta + cos theta = m and sec theta + cosec theta = n , prove that n(m² – 1) = 2m. About the author Ayla
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 , ⠀⠀⠀⠀⠀⠀▪︎⠀[tex]\sf \: sin \:\theta \: + \: \: cos \:\theta \: =\: m \:\\[/tex] ⠀⠀⠀⠀⠀⠀▪︎⠀[tex]\sf \: sec \:\theta \: + \: \: cosec \:\theta \: =\: n \:\\[/tex] ⠀⠀⠀Need To Proove : ⠀⠀⠀⠀⠀⠀▪︎⠀[tex]\bf{ \: n \:\bigg( \: m^2\: – \:1 \:\bigg) \: =\: 2m \: } \:\\[/tex] ⠀⠀Here , ⠀⠀⠀⠀⠀⠀[tex]\leadsto \: \bf L.H.S \:\:\sf =\: \: n \:\bigg( \: m^2\: – \:1 \:\bigg) \:\\[/tex] ⠀⠀⠀⠀⠀⠀[tex]\leadsto \: \bf R.H.S \:\:\sf =\: \: 2m \:\\[/tex] [tex]\qquad \bigstar \underline {\boldsymbol {Now \: by \:taking \: the \: \:L.H.S \:\: :}}\\[/tex] ⠀⠀⠀⠀⠀⠀[tex]\leadsto \: \bf L.H.S \:\:\sf =\: \: n \:\bigg( \: m^2\: – \:1 \:\bigg) \:\\\\[/tex] [tex]\qquad \dashrightarrow \:\sf \: \: n \:\bigg( \: m^2\: – \:1 \:\bigg) \:\\[/tex] ⠀⠀⠀⠀⠀⠀[tex]\underline {\boldsymbol{\star\:Now \: By \: Substituting \: the \: given \: Values \::}}\\[/tex] [tex]\qquad \dashrightarrow \:\sf \: \: n \:\bigg( \: m^2\: – \:1 \:\bigg) \:\\[/tex] ⠀⠀⠀▪︎⠀[tex]\sf \: sin \:\theta \: + \: \: cos \:\theta \: =\: m \:\\[/tex] ⠀⠀⠀▪︎⠀[tex]\sf \: sec \:\theta \: + \: \: cosec \:\theta \: =\: n \:\\[/tex] [tex]\qquad \dashrightarrow \:\sf \: \: n \:\bigg( \: m^2\: – \:1 \:\bigg) \:\\[/tex] [tex]\qquad \dashrightarrow \:\sf \: \bigg(\:sec \:\theta \: + \: \: cosec \:\theta \: \bigg) \: \:\bigg[ \: \bigg(\:sin \:\theta \: + \: \: cos \:\theta \: \bigg)^2\: – \:1 \:\bigg] \:\\\\[/tex] [tex]\qquad \because\:\:\bigg\lgroup \sf{ \red{ sec \: \theta \: =\:\dfrac{1}{\:cos \:\theta }}}\bigg\rgroup \\\\[/tex] ⠀⠀⠀⠀⠀AND , [tex]\qquad \because\:\:\bigg\lgroup \sf{ \red{ cosec \: \theta \: =\:\dfrac{1}{\:sin \:\theta }}}\bigg\rgroup \\\\[/tex] [tex]\qquad \dashrightarrow \:\sf \: \bigg(\:sec \:\theta \: + \: \: cosec \:\theta \: \bigg) \: \:\bigg[ \: \bigg(\:sin \:\theta \: + \: \: cos \:\theta \: \bigg)^2\: – \:1 \:\bigg] \:\\\\[/tex] [tex]\qquad \dashrightarrow \:\sf \: \bigg(\:\dfrac{1}{cos \:\theta } \: + \: \: \dfrac{1}{cos \:\theta} \: \bigg) \: \:\bigg[ \: \bigg(\:sin \:\theta \: + \: \: cos \:\theta \: \bigg)^2\: – \:1 \:\bigg] \:\\\\[/tex] [tex]\qquad \dashrightarrow \:\sf \: \bigg(\:\dfrac{\: sin \:\theta \: + \: \: cos \:\theta \: }{cos \:\theta\:\sin \:\theta } \: \: \bigg) \: \:\bigg[ \: \bigg(\:sin \:\theta \: + \: \: cos \:\theta \: \bigg)^2\: – \:1 \:\bigg] \:\\\\[/tex] [tex]\qquad \because\:\:\bigg\lgroup \sf{ \red{ ( \: a +\:b)^2\:\:=\:a^2 + b^2 + 2ab }}\bigg\rgroup \\\\[/tex] [tex]\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] \:\\\\[/tex] [tex]\qquad \because\:\:\bigg\lgroup \sf{ \red{ sin^2 \: \theta \: +\:\:cos^2 \:\theta\:\:=\:1 }}\bigg\rgroup \\\\[/tex] [tex]\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] \:\\\\[/tex] [tex]\qquad \dashrightarrow \:\sf \: \bigg(\:\dfrac{\: sin \:\theta \: + \: \: cos \:\theta \: }{cos \:\theta\:\sin \:\theta } \: \: \bigg) \: \:\bigg[ \: \:1 \:+\: 2cos \:\theta\:\sin \:\theta \: – \:1 \:\bigg] \:\\\\[/tex] [tex]\qquad \dashrightarrow \:\sf \: \bigg(\:\dfrac{\: sin \:\theta \: + \: \: cos \:\theta \: }{cos \:\theta\:\sin \:\theta } \: \: \bigg) \: \:\bigg[ \: \cancel {\:1 }\:+\: 2cos \:\theta\:\sin \:\theta \:\cancel {- \:1 }\:\bigg] \:\\\\[/tex] [tex]\qquad \dashrightarrow \:\sf \: \bigg(\:\dfrac{\: sin \:\theta \: + \: \: cos \:\theta \: }{cos \:\theta\:\sin \:\theta } \: \: \bigg) \: \:\bigg[ \: \:\: 2\:cos \:\theta\:\sin \:\theta \: \:\bigg] \:\\\\[/tex] [tex]\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) \: \: \:\\\\[/tex] ⠀⠀⠀▪︎⠀[tex]\sf \: sin \:\theta \: + \: \: cos \:\theta \: =\: m \:\\\\[/tex] [tex]\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 \\[/tex] ⠀⠀⠀⠀⠀[tex]\leadsto \: \bf L.H.S \:\:\sf =\: \: 2m \:\\[/tex] ⠀⠀⠀⠀⠀[tex]\leadsto \: \bf R.H.S \:\:\sf =\: \: 2m \:\\[/tex] ⠀⠀As , We can see that here , [ L.H.S = R.H.S ] [tex]\therefore {\underline {\bf{ Hence, \:Proved\:}}}\\\\\\[/tex] ⠀⠀⠀⠀⠀━━━━━━━━━━━━━━━━━━━⠀ Reply
Missbelladonna answer is correct
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 ,
⠀⠀⠀⠀⠀⠀▪︎⠀[tex]\sf \: sin \:\theta \: + \: \: cos \:\theta \: =\: m \:\\[/tex]
⠀⠀⠀⠀⠀⠀▪︎⠀[tex]\sf \: sec \:\theta \: + \: \: cosec \:\theta \: =\: n \:\\[/tex]
⠀⠀⠀Need To Proove :
⠀⠀⠀⠀⠀⠀▪︎⠀[tex]\bf{ \: n \:\bigg( \: m^2\: – \:1 \:\bigg) \: =\: 2m \: } \:\\[/tex]
⠀⠀Here ,
⠀⠀⠀⠀⠀⠀[tex]\leadsto \: \bf L.H.S \:\:\sf =\: \: n \:\bigg( \: m^2\: – \:1 \:\bigg) \:\\[/tex]
⠀⠀⠀⠀⠀⠀[tex]\leadsto \: \bf R.H.S \:\:\sf =\: \: 2m \:\\[/tex]
[tex]\qquad \bigstar \underline {\boldsymbol {Now \: by \:taking \: the \: \:L.H.S \:\: :}}\\[/tex]
⠀⠀⠀⠀⠀⠀[tex]\leadsto \: \bf L.H.S \:\:\sf =\: \: n \:\bigg( \: m^2\: – \:1 \:\bigg) \:\\\\[/tex]
[tex]\qquad \dashrightarrow \:\sf \: \: n \:\bigg( \: m^2\: – \:1 \:\bigg) \:\\[/tex]
⠀⠀⠀⠀⠀⠀[tex]\underline {\boldsymbol{\star\:Now \: By \: Substituting \: the \: given \: Values \::}}\\[/tex]
[tex]\qquad \dashrightarrow \:\sf \: \: n \:\bigg( \: m^2\: – \:1 \:\bigg) \:\\[/tex]
⠀⠀⠀▪︎⠀[tex]\sf \: sin \:\theta \: + \: \: cos \:\theta \: =\: m \:\\[/tex]
⠀⠀⠀▪︎⠀[tex]\sf \: sec \:\theta \: + \: \: cosec \:\theta \: =\: n \:\\[/tex]
[tex]\qquad \dashrightarrow \:\sf \: \: n \:\bigg( \: m^2\: – \:1 \:\bigg) \:\\[/tex]
[tex]\qquad \dashrightarrow \:\sf \: \bigg(\:sec \:\theta \: + \: \: cosec \:\theta \: \bigg) \: \:\bigg[ \: \bigg(\:sin \:\theta \: + \: \: cos \:\theta \: \bigg)^2\: – \:1 \:\bigg] \:\\\\[/tex]
[tex]\qquad \because\:\:\bigg\lgroup \sf{ \red{ sec \: \theta \: =\:\dfrac{1}{\:cos \:\theta }}}\bigg\rgroup \\\\[/tex]
⠀⠀⠀⠀⠀AND ,
[tex]\qquad \because\:\:\bigg\lgroup \sf{ \red{ cosec \: \theta \: =\:\dfrac{1}{\:sin \:\theta }}}\bigg\rgroup \\\\[/tex]
[tex]\qquad \dashrightarrow \:\sf \: \bigg(\:sec \:\theta \: + \: \: cosec \:\theta \: \bigg) \: \:\bigg[ \: \bigg(\:sin \:\theta \: + \: \: cos \:\theta \: \bigg)^2\: – \:1 \:\bigg] \:\\\\[/tex]
[tex]\qquad \dashrightarrow \:\sf \: \bigg(\:\dfrac{1}{cos \:\theta } \: + \: \: \dfrac{1}{cos \:\theta} \: \bigg) \: \:\bigg[ \: \bigg(\:sin \:\theta \: + \: \: cos \:\theta \: \bigg)^2\: – \:1 \:\bigg] \:\\\\[/tex]
[tex]\qquad \dashrightarrow \:\sf \: \bigg(\:\dfrac{\: sin \:\theta \: + \: \: cos \:\theta \: }{cos \:\theta\:\sin \:\theta } \: \: \bigg) \: \:\bigg[ \: \bigg(\:sin \:\theta \: + \: \: cos \:\theta \: \bigg)^2\: – \:1 \:\bigg] \:\\\\[/tex]
[tex]\qquad \because\:\:\bigg\lgroup \sf{ \red{ ( \: a +\:b)^2\:\:=\:a^2 + b^2 + 2ab }}\bigg\rgroup \\\\[/tex]
[tex]\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] \:\\\\[/tex]
[tex]\qquad \because\:\:\bigg\lgroup \sf{ \red{ sin^2 \: \theta \: +\:\:cos^2 \:\theta\:\:=\:1 }}\bigg\rgroup \\\\[/tex]
[tex]\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] \:\\\\[/tex]
[tex]\qquad \dashrightarrow \:\sf \: \bigg(\:\dfrac{\: sin \:\theta \: + \: \: cos \:\theta \: }{cos \:\theta\:\sin \:\theta } \: \: \bigg) \: \:\bigg[ \: \:1 \:+\: 2cos \:\theta\:\sin \:\theta \: – \:1 \:\bigg] \:\\\\[/tex]
[tex]\qquad \dashrightarrow \:\sf \: \bigg(\:\dfrac{\: sin \:\theta \: + \: \: cos \:\theta \: }{cos \:\theta\:\sin \:\theta } \: \: \bigg) \: \:\bigg[ \: \cancel {\:1 }\:+\: 2cos \:\theta\:\sin \:\theta \:\cancel {- \:1 }\:\bigg] \:\\\\[/tex]
[tex]\qquad \dashrightarrow \:\sf \: \bigg(\:\dfrac{\: sin \:\theta \: + \: \: cos \:\theta \: }{cos \:\theta\:\sin \:\theta } \: \: \bigg) \: \:\bigg[ \: \:\: 2\:cos \:\theta\:\sin \:\theta \: \:\bigg] \:\\\\[/tex]
[tex]\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) \: \: \:\\\\[/tex]
⠀⠀⠀▪︎⠀[tex]\sf \: sin \:\theta \: + \: \: cos \:\theta \: =\: m \:\\\\[/tex]
[tex]\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 \\[/tex]
⠀⠀⠀⠀⠀[tex]\leadsto \: \bf L.H.S \:\:\sf =\: \: 2m \:\\[/tex]
⠀⠀⠀⠀⠀[tex]\leadsto \: \bf R.H.S \:\:\sf =\: \: 2m \:\\[/tex]
⠀⠀As , We can see that here , [ L.H.S = R.H.S ]
[tex]\therefore {\underline {\bf{ Hence, \:Proved\:}}}\\\\\\[/tex]
⠀⠀⠀⠀⠀━━━━━━━━━━━━━━━━━━━⠀