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| 1. |
Identify-(I)the substance oxidised and (ii) substance reduced in the give equation: H²S(g)+Br²(aq)—2HBr(aq)+S(s) |
| Answer» TION:Step-by-step EXPLANATION:★Given -\bf \rm{ \dfrac{sec \theta \: + TAN \theta}{sec \theta \: - tan \theta} } secθ−tanθsecθ+tanθ ★To prove -\bf\rm{ {\huge(} \dfrac{1 + sin \theta}{cos \theta} {\huge)}^{2} }( cosθ1+sinθ ) 2 ★Solution -L.H.S\longmapsto \bf \rm{ \dfrac{sec \theta \: + tan \theta}{sec \theta \: - tan \theta} }⟼ secθ−tanθsecθ+tanθ \longmapsto \bf \rm{ \dfrac{ \dfrac{1}{cos \theta} \: + \dfrac{sin \theta}{cos \theta} }{ \dfrac{1}{cos \theta} \: - \dfrac{sin \theta}{cos \theta} } }⟼ cosθ1 − cosθsinθ cosθ1 + cosθsinθ \bf \longmapsto \rm{ \dfrac{ \dfrac{1 + sin \theta}{cos \theta} }{ \dfrac{1 - sin \theta}{cos \theta} } }⟼ cosθ1−sinθ cosθ1+sinθ \longmapsto \bf \rm{ \dfrac{1 + sin \theta}{ \cancel{cos \theta}} \times \dfrac{ \cancel{cos \theta}}{ 1 - sin \theta} }⟼ cosθ 1+sinθ × 1−sinθcosθ \longmapsto \bf \rm{ \dfrac{1 + sin \theta}{1 - sin \theta}}⟼ 1−sinθ1+sinθ \longmapsto \bf \rm{ \dfrac{1 + sin \theta}{1 - sin \theta} \times \dfrac{1 + sin \theta}{1 + sin \theta} }⟼ 1−sinθ1+sinθ × 1+sinθ1+sinθ \longmapsto \bf \rm{ \dfrac{(1 + sin \theta)^{2} }{ {1}^{2} - sin^{2} \theta}}⟼ 1 2 −sin 2 θ(1+sinθ) 2 \longmapsto \bf \rm{ \dfrac{(1 + sin \theta)^{2} }{ {1} - sin^{2} \theta}}⟼ 1−sin 2 θ(1+sinθ) 2 ∵ sin²∅ + cos ²∅ = 1=> cos²∅ = 1 - sin²∅\longmapsto \bf \rm{ \dfrac{(1 + sin \theta)^{2} }{ {cos}^{2} \theta}}⟼ cos 2 θ(1+sinθ) 2 \longmapsto \bf \rm{ \dfrac{(1 + sin \theta)^{2} }{ ( {cos} \theta) ^{2} }}⟼ (cosθ) 2 (1+sinθ) 2 \longmapsto \bf \GREEN{\rm{ { {{\huge(} \dfrac{1 + sin \theta}{cos \theta} {\huge)}^{2} }}}= R.H.S}⟼( cosθ1+sinθ ) 2 =R.H.S★More to know -\rm{sin \theta = \dfrac{1}{cosec \theta} }sinθ= cosecθ1 \rm{cos \theta = \dfrac{1}{sec \theta} }cosθ= secθ1 \rm{tan \theta = \dfrac{1}{cot \theta} }tanθ= cotθ1 \rm{cosec \theta = \dfrac{1}{sin \theta} }cosecθ= sinθ1 \rm{sec \theta = \dfrac{1 }{cos \theta} }secθ= cosθ1 \rm{cot \theta = \dfrac{1 }{tan \theta} }cotθ= tanθ1 | |