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Trigonometry 18 Online
OpenStudy (anonymous):

prove that : root of sec^2 A +cosec ^2 A= tan A+cot A

OpenStudy (sidsiddhartha):

use \[secA=1/cosA\\cosecA=1/sinA\\then \\ LHS=\sqrt{\sec^2A+cosec^2A}=\sqrt{\frac{ 1 }{ \cos^2A }+\frac{ 1 }{ \sin^2A }}=\sqrt{\frac{ \sin^2A+\cos^2A }{ \sin^2A.\cos^2A }}\\=\frac{ 1 }{ sinA.cosA }\]

OpenStudy (sidsiddhartha):

now for the RHS \[tanA+cotA=\frac{ sinA }{ cosA }+\frac{ cosA }{ sinA }=\frac{ \sin^2A+\cos^2A }{ sinA.cosA }=\frac{ 1 }{ sinA.cosA }=LHS\]

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