tan2θ(1+tan2θ)+cot2θ(1+cot2θ)=1
To prove, tan2 θ1+tan2 θ+cot2 θ1+cot2 θ=1LHS=tan2 θ1+tan2 θ+cot2 θ1+cot2 θ=tan2 θsec2 θ+cot2 θcosec2 θ=1sec2 θ×sin2 θcos2 θ+1cosec2 θ×cos2 θsin2 θcos2 θ×sin2 θcos2 θ+sin2 θ×cos2 θsin2 θ=sin2 θ+cos2 θ=1=RHS
(i) sin2θ+1(1+tan2θ)=1 (ii) 11+tan2θ+11+cot2θ=1
(i) (sec2θ−1)cot2θ=1 (ii) (sec2θ−1)(cosec2θ−1)=1
(iii) (1−cos2θ)sec2θ=tan2θ
Prove:
(i)sin2θ+11+tan2θ=1
(ii)11+tan2θ+11+cot2θ=1