Question

cosecA-secA=(root5)/2 THEN VALUE of cosA - sinA

Answer 1

1/sinA-1/cosA=√5/2.

(cosA-sinA)/sinAcosA=√5/2. Remember this for later.

Squaring both sides:

(cos^2(A)+sin^2(A)-2sinAcosA)/sin^2(A)cos^2(A)=5/4;

(1-2sinAcosA)/sin^2(A)cos^2(A)=5/4.

Let x=sinAcosA:

(1-2x)/x^2=5/4; 4(1-2x)=5x^2; 5x^2+8x-4=0=(5x-2)(x+2),

so x=sinAcosA=2/5 or -2.

But sinAcosA cannot =-2 for any A because sinAcosA=sin2A/2 and sin2A can never=-4 (sine is in range -1 to +1). Therefore sinAcosA=2/5 and, as we saw earlier, cosA-sinA=(√5/2)*sinAcosA=(√5/2)(2/5)=1/√5 or √5/5.