If $\cos \left(2 \theta+54^{\circ}\right)=\sin \theta, 0^{\circ}<\left(2 \theta+54^{\circ}\right)<90^{\circ}$, then what is the value of $\frac{1}{\cot 5 \theta+\sec \frac{5 \theta}{2}} ?$
Answer & explanation
Correct answer: option 3
We are given :-
cos (2θ + 54º) = sinθ
{ using, Iff A + B = 90º , then sinA = cosB }
So, 2θ + 54º + θ= 90º
3θ = 36º
θ = 12º
Now,
\(\frac{1}{ cot5θ + sec5θ/2}\)
= \(\frac{1}{ cot60º + sec30º }\)
= \(\frac{1}{ 1/√3 + 2/√3 }\)
= \(\frac{√3}{ 3 }\)