Question

1-sinA+cosA/SinA+CosA-1=1+CosA/SinA

Answer 1

We know that sec A = 1/cosA and if you take cosA outside,

LHS becomes cosA( 1/cosA - sinA/cosA + cosA/cosA) / cosA(sinA/cosA +cosA/cosA-1/cosA)

LHS= cosA(secA-tanA+1) /cosA(tanA+1-secA)

= (secA-tanA+1)/(tanA+1-secA)

= (secA -tanA + (sec sq. A - tan sq.A)) /

(tanA+1-secA)

(By using the formula sec sq. A - tan sq. A = 1= (secA+tanA) (secA - tanA)

LHS becomes = secA + tanA

=( 1/cosA) + (cosA/sinA)

=(1 + cosA) /sinA= RHS