Math, asked by jollysoham, 8 months ago

1+tanA/sinA+1+cotA/cosA=2(secA+cosecA)

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Answered by tanmayee28

Answer:

See the image , hope it will help u.

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Answered by sandy1816

$\frac{1 + cotA}{cosA} + \frac{1 + tanA}{sinA} \\ \\ = \frac{sinA + cosA}{sinA \: cosA} + \frac{cosA + sinA}{sinA \: cosA} \\ \\ = (sinA + cosA)( \frac{2}{sinA \: cosA} )$

$= 2( \frac{sinA + cosA}{sinA \: cosA} ) \\ \\ = 2(secA + cosecA)$

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1+tanA/sinA+1+cotA/cosA=2(secA+cosecA)​

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1+tanA/sinA+1+cotA/cosA=2(secA+cosecA)