If tan2(θ)=2tan(ϕ)+1, then cos(2θ)+sin2(ϕ) equals
-1
0
1
none of these
Explanation for correct options
Given: tan2(θ)=2tan(ϕ)+1 ---------(1)
we know that cos(2θ)=1-tan2(θ)1+tan2(θ)
⇒cos(2θ)=1-2tan2(ϕ)+11+2tan2(ϕ)+1∵tan2(θ)=2tan(ϕ)+1⇒cos(2θ)=1-2tan2(ϕ)-11+2tan2(ϕ)+1⇒cos(2θ)=-2tan2(ϕ)2+2tan2(ϕ)⇒cos(2θ)=-tan2(ϕ)sec2ϕ1+tan2(x)=sec2x⇒cos(2θ)=-tan2(ϕ)sec2ϕ⇒cos(2θ)=-sin2(ϕ)cos2(ϕ)1cos2(ϕ)⇒cos(2θ)=-sin2(ϕ)------(2)
From (2)
cos(2θ)+sin2(ϕ)=-sin2(ϕ)+sin2(ϕ)⇒=0
Hence, option B is correct.
If x<(-2), then1-1+x equals to: