KeerthanaPosted on (1+tan2 A)cotAcosec2 A\frac{\left(1+\tan ^{2} \mathrm{~A}\right) \cot \mathrm{A}}{\operatorname{cosec}^{2} \mathrm{~A}}cosec2 A(1+tan2 A)cotA is equal to acot A btan A csin A dcos A Answer : Option BExplanation : Expression: =(1+tan2 A)cotAcosec2 A=\frac{\left(1+\tan ^{2} \mathrm{~A}\right) \cot \mathrm{A}}{\operatorname{cosec}^{2} \mathrm{~A}}=cosec2 A(1+tan2 A)cotA =sec2 A⋅cotA⋅sin2 A=\sec ^{2} \mathrm{~A} \cdot \cot \mathrm{A} \cdot \sin ^{2} \mathrm{~A}=sec2 A⋅cotA⋅sin2 A (sinA⋅cosecA=1)(\sin A \cdot \operatorname{cosec} A=1)(sinA⋅cosecA=1) =1cos2 A⋅cosAsinA⋅sin2 A=\frac{1}{\cos ^{2} \mathrm{~A}} \cdot \frac{\cos \mathrm{A}}{\sin \mathrm{A}} \cdot \sin ^{2} \mathrm{~A}=cos2 A1⋅sinAcosA⋅sin2 A =sinAcosA=tanA=\frac{\sin \mathrm{A}}{\cos \mathrm{A}}=\tan \mathrm{A}=cosAsinA=tanA Rate This:NaN / 5 - 1 votesAdd comment