If secθ+tanθ=y,secθ+tanθ=y, simplify y2−1y2+1 in terms of θ.
Answer:
sinθ
- y2−1y2+1=(secθ+tanθ)2−1(secθ+tanθ)2+1=sec2θ+tan2θ+2tanθsecθ−1sec2θ+tan2θ+2tanθsecθ+1=(sec2θ−1)+tan2θ+2tanθsecθsec2θ+(1+tan2θ)+2tanθsecθ=2tan2θ+2tanθsecθ2sec2θ+2tanθsecθ=2tanθ(tanθ+secθ)2secθ(tanθ+secθ)=sinθ