If `a^2+b^2+2a bcostheta=1,c^2+a^2+2c dcostheta=1` and `a c+b d+(a d+b c)costheta=0,` then prove that `a^2+c^2=cos e c^2theta`

Explanation

`a^2+b^2+2abcostheta=1`
`rArr(b+acos theta)^2=1-a^2sin^2theta`
`c^2+d^2+2cdcostheta=1`
`rArr(d+c costheta)^2=1-c^2sin^2theta`
`ac+bd+(ad+bc)costheta=0`
`rArr (b+acostheta)(d+c costheta)=-ac sin^2theta`
`rArr(b+acostheta)^2(d+c costheta)^2=(-ac sin^2theta)^2`
`rArr (1-a^2sin^2theta)(1-c^2sin^2theta)=a^2c^2sin^4theta`
`rArr (a^2+c^2)sin^2theta=1`
`rArr (a^2+c^2)=cosec^2theta`