If xcosθ=ycosθ+2π3=zcosθ+4π3, then the value of 1x+1y+1z is equal to
1
2
0
3cosθ
Explanation for the correct option.
Let xcosθ=ycosθ+2π3=zcosθ+4π3=k
Now,
kx=cosθ.....1ky=cosθ+2π3......2kz=cosθ+4π3......3
By adding 1,2,and3 we get
kx+ky+kz=cosθ+cosθ+2π3+cosθ+4π3⇒k1x+1y+1z=cosθ+cosθ+2π3+cosθ+4π3=2cos2θ+4π32cos-4π32+cosθ+2π3[∵cosA+cosB=2cosA+B2cosA-B2]=-2cosθ+2π3cos2π3+cosθ+2π3=-2×12cosθ+2π3+cosθ+2π3bycos2π3=12=0
Hence, option C is correct.
{1(sec2θ−cos2θ)+1(cosec2θ−sin2θ)}(sin2θcos2θ)=1−sin2θcos2θ2+sin2θcos2θ