Prove that tanθ1cotθ+cotθ1tanθ=1+secθcosecθ.
 
L.H.S \(=\) tanθ1cotθ+cotθ1tanθ
 
 Now, simplyfing this then we get, 
 
 \(LCM=\)
 
=tan3θ1tanθtanθ1
 
By applying the expansion of \(a^3-b^3\) then we get, 
 
\(LHS=\)
 
By applying \(tan \theta= \frac{sin \theta}{cos \theta}\) , \(cot \theta = \frac{cos \theta}{sin \theta}\) and simplyfing then we get, 
 
\(LHS=\)
 
Now, simplifying this then we get, 
 
\(=\) \(1 + sec \theta \cdot cosec \theta\)
 
\(=\) R.H.S
 
Hence proved.
Answer variants:
=tan2θtanθ11tanθtanθ1
=tanθ+1tan2θ+tanθ+1tanθtanθ1
=tanθ1tan2θ+tanθ+1tanθtanθ1
=1+sin2θ+cos2θcosθsinθ
=tan2θtanθ+11tanθtanθ+1