prove that tanA+secA-1/tanA-secA+1

get this please ..............
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1 Answer

This can be written:

(sinA/cosA+1/cosA-1)/(sinA/cosA-1/cosA+1). Multiply top and bottom by cosA:

(sinA+1-cosA)/(sinA-1+cosA).

Multiply top and bottom by sinA+1-cosA:

(sinA+1-cosA)2/(sin2A-(1-cosA)2)=

(sin2A+1-2cosA+cos2A+2sinA-2sinAcosA)/(sin2A-1+2cosA-cos2A)=

(2-2cosA+2sinA(1-cosA))/(2cosA-2cos2A). Divide top and bottom by 2 and factorise:

(1-cosA)(1+sinA)/(cosA(1-cosA))=(1+sinA)/cosA=tanA+secA.

by Top Rated User (1.2m points)

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