If sin theta + cosec theta =2 then find sin^19theta +cosec^19theta =?? PLZ HELP CORRECT ANSWER WILL BE MARKED AS BRAINLIEST About the author Savannah
Appropriate Question : If [tex] \sin \theta + \cosec \theta = 2 \\[/tex] . Then , Find the value of [tex] \sin^{19}\theta + \cosec ^{19} \theta \\ [/tex] . ⠀⠀⠀⠀⠀━━━━━━━━━━━━━━━━━━━⠀ Given : [tex] \sin \theta + \cosec \theta = 2 \\[/tex] Exigency to find : The Value of : [tex] \sin^{19}\theta + \cosec ^{19} \theta \\ [/tex] ⠀⠀⠀⠀⠀━━━━━━━━━━━━━━━━━━━⠀ [tex]\qquad \dag\:\:\bigg\lgroup \sf{ \sin \theta + \cosec \theta = 2 }\bigg\rgroup \\\\[/tex] ━━━━ Finding Value of : [tex] \sin\theta \:\:\&\:\: \cosec \theta \\ [/tex] Now , By Solving the Given : [tex]\qquad \longmapsto \sf \sin \theta + \cosec \theta = 2 \\\\ [/tex] [tex]\dag\:\:\it{ As,\:We\:know\:that\::}\\[/tex] [tex] \sf \cosec \theta = \dfrac{1}{\sin\theta } \\\\[/tex] [tex]\qquad \longmapsto \sf \sin \theta + \cosec \theta = 2 \\\\ [/tex] [tex]\qquad \longmapsto \sf \sin \theta + \dfrac{1}{\sin\theta } = 2 \\\\ [/tex] [tex]\qquad \longmapsto \sf \sin \theta + \dfrac{1}{\sin\theta } – 2 = 0 \\\\ [/tex] [tex]\qquad \longmapsto \sf \dfrac{\sin^2 \theta + 1 – 2\sin\theta}{\sin\theta } = 0 \\\\ [/tex] [tex]\qquad \longmapsto \sf \sin^2 \theta + 1 – 2\sin\theta = 0 \times \sin \theta \\\\ [/tex] [tex]\qquad \longmapsto \sf \sin^2 \theta + 1 – 2\sin\theta = 0 \\\\ [/tex] [tex]\dag\:\:\it{ As,\:We\:know\:that\::}\\[/tex] [tex] \sf (a – b)^2 = a^2 + b^2 – 2ab \\\\[/tex] [tex]\qquad \longmapsto \sf \sin^2 \theta + 1 – 2\sin\theta = 0 \\\\ [/tex] [tex]\qquad \longmapsto \sf (\sin \theta – 1 )^2 = 0 \\\\ [/tex] [tex]\qquad \longmapsto \sf \sin \theta – 1 = 0 \\\\ [/tex] [tex]\qquad \longmapsto \frak{\underline{\purple{\:\sin\theta = 1 }} }\bigstar \\[/tex] [tex]\dag\:\:\it{ As,\:We\:know\:that\::}\\[/tex] [tex] \sf \cosec \theta = \dfrac{1}{\sin\theta } \\\\[/tex] ⠀⠀⠀⠀⠀Here, [tex]\sin \theta = 1 [/tex] ⠀⠀⠀⠀⠀⠀[tex]\underline {\boldsymbol{\star\:Now \: By \: Substituting \: the \: known \: Values \::}}\\[/tex] [tex]\qquad \longmapsto \sf \cosec \theta = \dfrac{1}{\sin\theta } \\\\[/tex] [tex]\qquad \longmapsto \sf \cosec \theta = \cancel{\dfrac{1}{1 }} \\\\[/tex] [tex]\qquad \longmapsto \frak{\underline{\purple{\:\cosec\theta = 1 }} }\bigstar \\[/tex] ━━━━ Finding Value of : [tex] \sin^{19}\theta + \cosec ^{19} \theta \\ [/tex] ⠀⠀⠀⠀⠀Here [tex]\sin \theta = 1 \:\:\:\& \:\:\:\: \cosec \theta = 1 [/tex] ⠀⠀⠀⠀⠀⠀[tex]\underline {\boldsymbol{\star\:Now \: By \: Substituting \: the \: known \: Values \::}}\\[/tex] [tex]\qquad \longmapsto \sf \sin^{19}\theta + \cosec ^{19} \theta \\ [/tex] [tex]\qquad \longmapsto \sf 1^{19} + 1 ^{19} \\ [/tex] [tex]\qquad \longmapsto \sf 1 + 1 \\ [/tex] [tex]\qquad \longmapsto \bf \bigg( \:\:2 \:\:\bigg)\qquad \longrightarrow \:\: Required \:AnswEr \:\\ [/tex] Therefore, ⠀⠀⠀⠀⠀[tex]\therefore {\underline{ \mathrm {\:The\:value \:of\: \sin^{19}\theta + \cosec ^{19} \theta \:is\:\bf{2}}}}\\[/tex] ⠀⠀⠀⠀⠀━━━━━━━━━━━━━━━━━━━⠀ Reply
Appropriate Question :
⠀⠀⠀⠀⠀━━━━━━━━━━━━━━━━━━━⠀
Given : [tex] \sin \theta + \cosec \theta = 2 \\[/tex]
Exigency to find : The Value of : [tex] \sin^{19}\theta + \cosec ^{19} \theta \\ [/tex]
⠀⠀⠀⠀⠀━━━━━━━━━━━━━━━━━━━⠀
[tex]\qquad \dag\:\:\bigg\lgroup \sf{ \sin \theta + \cosec \theta = 2 }\bigg\rgroup \\\\[/tex]
━━━━ Finding Value of : [tex] \sin\theta \:\:\&\:\: \cosec \theta \\ [/tex]
Now , By Solving the Given :
[tex]\qquad \longmapsto \sf \sin \theta + \cosec \theta = 2 \\\\ [/tex]
[tex]\dag\:\:\it{ As,\:We\:know\:that\::}\\[/tex]
[tex]\qquad \longmapsto \sf \sin \theta + \cosec \theta = 2 \\\\ [/tex]
[tex]\qquad \longmapsto \sf \sin \theta + \dfrac{1}{\sin\theta } = 2 \\\\ [/tex]
[tex]\qquad \longmapsto \sf \sin \theta + \dfrac{1}{\sin\theta } – 2 = 0 \\\\ [/tex]
[tex]\qquad \longmapsto \sf \dfrac{\sin^2 \theta + 1 – 2\sin\theta}{\sin\theta } = 0 \\\\ [/tex]
[tex]\qquad \longmapsto \sf \sin^2 \theta + 1 – 2\sin\theta = 0 \times \sin \theta \\\\ [/tex]
[tex]\qquad \longmapsto \sf \sin^2 \theta + 1 – 2\sin\theta = 0 \\\\ [/tex]
[tex]\dag\:\:\it{ As,\:We\:know\:that\::}\\[/tex]
[tex]\qquad \longmapsto \sf \sin^2 \theta + 1 – 2\sin\theta = 0 \\\\ [/tex]
[tex]\qquad \longmapsto \sf (\sin \theta – 1 )^2 = 0 \\\\ [/tex]
[tex]\qquad \longmapsto \sf \sin \theta – 1 = 0 \\\\ [/tex]
[tex]\qquad \longmapsto \frak{\underline{\purple{\:\sin\theta = 1 }} }\bigstar \\[/tex]
[tex]\dag\:\:\it{ As,\:We\:know\:that\::}\\[/tex]
⠀⠀⠀⠀⠀Here, [tex]\sin \theta = 1 [/tex]
⠀⠀⠀⠀⠀⠀[tex]\underline {\boldsymbol{\star\:Now \: By \: Substituting \: the \: known \: Values \::}}\\[/tex]
[tex]\qquad \longmapsto \sf \cosec \theta = \dfrac{1}{\sin\theta } \\\\[/tex]
[tex]\qquad \longmapsto \sf \cosec \theta = \cancel{\dfrac{1}{1 }} \\\\[/tex]
[tex]\qquad \longmapsto \frak{\underline{\purple{\:\cosec\theta = 1 }} }\bigstar \\[/tex]
━━━━ Finding Value of : [tex] \sin^{19}\theta + \cosec ^{19} \theta \\ [/tex]
⠀⠀⠀⠀⠀Here [tex]\sin \theta = 1 \:\:\:\& \:\:\:\: \cosec \theta = 1 [/tex]
⠀⠀⠀⠀⠀⠀[tex]\underline {\boldsymbol{\star\:Now \: By \: Substituting \: the \: known \: Values \::}}\\[/tex]
[tex]\qquad \longmapsto \sf \sin^{19}\theta + \cosec ^{19} \theta \\ [/tex]
[tex]\qquad \longmapsto \sf 1^{19} + 1 ^{19} \\ [/tex]
[tex]\qquad \longmapsto \sf 1 + 1 \\ [/tex]
[tex]\qquad \longmapsto \bf \bigg( \:\:2 \:\:\bigg)\qquad \longrightarrow \:\: Required \:AnswEr \:\\ [/tex]
Therefore,
⠀⠀⠀⠀⠀[tex]\therefore {\underline{ \mathrm {\:The\:value \:of\: \sin^{19}\theta + \cosec ^{19} \theta \:is\:\bf{2}}}}\\[/tex]
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