Question

please answer the questions in the image

Answer 1

Answer:

Oiii first i dont see the image ok what type of image is this

Answer 2

Consider the L.H.S

Consider the L.H.Ssinθ+cosθ−1sinθ−cosθ+1

Consider the L.H.Ssinθ+cosθ−1sinθ−cosθ+1=(sinθ+cosθ−1sinθ−cosθ+1)×(sinθ+cosθ+1sinθ+cosθ+1)

Consider the L.H.Ssinθ+cosθ−1sinθ−cosθ+1=(sinθ+cosθ−1sinθ−cosθ+1)×(sinθ+cosθ+1sinθ+cosθ+1)=(sinθ+cosθ−1sinθ+1−cosθ)×(sinθ+cosθ+1sinθ+1+cosθ)

Consider the L.H.Ssinθ+cosθ−1sinθ−cosθ+1=(sinθ+cosθ−1sinθ−cosθ+1)×(sinθ+cosθ+1sinθ+cosθ+1)=(sinθ+cosθ−1sinθ+1−cosθ)×(sinθ+cosθ+1sinθ+1+cosθ)=(sinθ+cosθ)2−12(sinθ+1)2−cos2θ

Consider the L.H.Ssinθ+cosθ−1sinθ−cosθ+1=(sinθ+cosθ−1sinθ−cosθ+1)×(sinθ+cosθ+1sinθ+cosθ+1)=(sinθ+cosθ−1sinθ+1−cosθ)×(sinθ+cosθ+1sinθ+1+cosθ)=(sinθ+cosθ)2−12(sinθ+1)2−cos2θ=sin2θ+cos2θ+2sinθcosθ−1sin2θ+1+2sinθ−cos2θ

Consider the L.H.Ssinθ+cosθ−1sinθ−cosθ+1=(sinθ+cosθ−1sinθ−cosθ+1)×(sinθ+cosθ+1sinθ+cosθ+1)=(sinθ+cosθ−1sinθ+1−cosθ)×(sinθ+cosθ+1sinθ+1+cosθ)=(sinθ+cosθ)2−12(sinθ+1)2−cos2θ=sin2θ+cos2θ+2sinθcosθ−1sin2θ+1+2sinθ−cos2θ 

Consider the L.H.Ssinθ+cosθ−1sinθ−cosθ+1=(sinθ+cosθ−1sinθ−cosθ+1)×(sinθ+cosθ+1sinθ+cosθ+1)=(sinθ+cosθ−1sinθ+1−cosθ)×(sinθ+cosθ+1sinθ+1+cosθ)=(sinθ+cosθ)2−12(sinθ+1)2−cos2θ=sin2θ+cos2θ+2sinθcosθ−1sin2θ+1+2sinθ−cos2θ Since, sin2θ+cos2θ=1

Consider the L.H.Ssinθ+cosθ−1sinθ−cosθ+1=(sinθ+cosθ−1sinθ−cosθ+1)×(sinθ+cosθ+1sinθ+cosθ+1)=(sinθ+cosθ−1sinθ+1−cosθ)×(sinθ+cosθ+1sinθ+1+cosθ)=(sinθ+cosθ)2−12(sinθ+1)2−cos2θ=sin2θ+cos2θ+2sinθcosθ−1sin2θ+1+2sinθ−cos2θ Since, sin2θ+cos2θ=1 

Consider the L.H.Ssinθ+cosθ−1sinθ−cosθ+1=(sinθ+cosθ−1sinθ−cosθ+1)×(sinθ+cosθ+1sinθ+cosθ+1)=(sinθ+cosθ−1sinθ+1−cosθ)×(sinθ+cosθ+1sinθ+1+cosθ)=(sinθ+cosθ)2−12(sinθ+1)2−cos2θ=sin2θ+cos2θ+2sinθcosθ−1sin2θ+1+2sinθ−cos2θ Since, sin2θ+cos2θ=1 Therefore,

Consider the L.H.Ssinθ+cosθ−1sinθ−cosθ+1=(sinθ+cosθ−1sinθ−cosθ+1)×(sinθ+cosθ+1sinθ+cosθ+1)=(sinθ+cosθ−1sinθ+1−cosθ)×(sinθ+cosθ+1sinθ+1+cosθ)=(sinθ+cosθ)2−12(sinθ+1)2−cos2θ=sin2θ+cos2θ+2sinθcosθ−1sin2θ+1+2sinθ−cos2θ Since, sin2θ+cos2θ=1 Therefore,=1+2sinθcosθ−11−cos2θ