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

Question

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

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

Answers ( )

    0
    2021-06-30T16:08:49+00:00

    Missbelladonna answer is correct

    0
    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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