If secA-tanA=1/2 and 0 less than A less than 90 then show that secA=5/4

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If secA-tanA=1/2 and 0 less than A less than 90 then show that secA=5/4

Consider secA-tanA=1/2 -(I)

[sec^2A-tan^2A=1]

⇒secA-tanA=(sec^2A-tan^2A)/2

⇒secA-tanA=(secA-tanA)(secA+tanA)/2

⇒1=(secA+tanA)/2

⇒secA+tanA=2-(II)

(I)⇒secA-tanA=1/2

(II)⇒secA+tanA=2

On solving (I) & (II)

⇒2secA=1/2+2

⇒2secA=5/2

⇒secA=5/4

Hence If secA-tanA=1/2 and 0 less than A less than 90 then secA=5/4 is proved.

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