$\frac{\cot^3 \theta}{cosec^2 \theta} + \frac{\tan^3 \theta}{\sec^2 \theta} + 2\sin \theta \cos \theta = ?$

cosecθ secθ

\(\frac{ cot²θ }{ cosec²θ }\) + \(\frac{ tan²θ }{ sec²θ }\) + 2sinθ.cosθ

= cos²θ + sin²θ + 2sinθ.cosθ 

= ( sinθ + cosθ )²

Put θ = 45º

= ( sin45º + cos45º )²

= ( \(\frac{1}{√2}\) + \(\frac{1}{√2}\) )²

= ( \(\frac{2}{√2}\) )²

= 2

Now, lets solve one by one for each option,

option 1 :-

cosecθ secθ

= cosec 45º .  sec 45º

= (√2). (√2)

= 2

( satisfied )

So, Ans :-  cosecθ secθ