\(\sqrt3\)1 (3-cot30^o)=tan³ 60^o-2sin60^o

tan θ+tan(90^o-θ)=secθ sec(90^o-θ)

If \(\sqrt3\) tan θ=1, then find the value of sin²θ-cos²θ

Show that \(\\ \frac { { cos }^{ 2 }({ 45 }^{ 0 }+\theta ){ +cos }^{ 2 }({ 45 }^{ 0 }-\theta ) }{ tan({ 60 }^{ 0 }+\theta )tan({ 30 }^{ 0 }-\theta ) } =1\)

Show that tan⁴θ+tan²θ=sec⁴θ-sec²θ.