Question

sin (90-A) /cos A+tanA/cot(90-A) +sec(90-A)sinA=

Answer 1

$= > \frac{SIN(90 - A)}{COSA} + \frac{TANA}{COT(90 - A)} + \frac{SEC(90 - A)}{1} \times SINA \\ \\ = > \frac{COSA}{COSA} + \frac{TANA}{TANA} + COSECA \times SINA \\ \\ = > 1 + 1 + \frac{1}{SINA} \times SINA \\ \\ = > 1 + 1 + 1 = 3$
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Answer 2

Hi ,

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We know that ,

i ) sin( 90 - A ) = cosA

ii ) cot( 90 - A ) = tanA

iii ) sec( 90 - A ) = cosecA

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sin(90-A)/cosA + tanA/cot(90-A)

+secA(90-A)sinA

= ( cosA/cosA )+(tanA/tanA)+cosecAsinA

= 1 + 1 + (1/sinA)sinA

= 2 + 1

= 3

I hope this helps you.

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