$\frac{\cot^3 \theta}{cosec^2 \theta} + \frac{\tan^3 \theta}{\sec^2 \theta} + 2\sin \theta \cos \theta = ?$
Answer & explanation
Correct answer: option 1
\(\frac{ cot²θ }{ cosec²θ }\) + \(\frac{ tan²θ }{ sec²θ }\) + 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