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Question
Show that : `sinA/sin(90^circ - A) + cosA/cos(90^circ - A) = sec A cosec A`
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Solution
L.H.S. = `sinA/(sin (90^circ - A)) + cosA/(cos(90^circ - A))`
= `sinA/cosA + cosA/sinA`
= `(sin^2A + cos^2A)/(cosAsinA)` ...(∵ sin2 A + cos2 A = 1)
= `1/(cosAsinA)`
= sec A cosec A = R.H.S.
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