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(1+tan2 A)cotAcosec2 A\frac{\left(1+\tan ^{2} \mathrm{~A}\right) \cot \mathrm{A}}{\operatorname{cosec}^{2} \mathrm{~A}} is equal to

a

cot A

b

tan A

c

sin A

d

cos A

Answer : Option B
Explanation :

Expression:

=(1+tan2 A)cotAcosec2 A=\frac{\left(1+\tan ^{2} \mathrm{~A}\right) \cot \mathrm{A}}{\operatorname{cosec}^{2} \mathrm{~A}}

=sec2 AcotAsin2 A=\sec ^{2} \mathrm{~A} \cdot \cot \mathrm{A} \cdot \sin ^{2} \mathrm{~A}

(sinAcosecA=1)(\sin A \cdot \operatorname{cosec} A=1)

=1cos2 AcosAsinAsin2 A=\frac{1}{\cos ^{2} \mathrm{~A}} \cdot \frac{\cos \mathrm{A}}{\sin \mathrm{A}} \cdot \sin ^{2} \mathrm{~A}

=sinAcosA=tanA=\frac{\sin \mathrm{A}}{\cos \mathrm{A}}=\tan \mathrm{A}

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