If $7 \cos ^2 \theta+5 \sin ^2 \theta-6=0,\left(0^{\circ}<\theta<90^{\circ}\right)$, then what is the value of $\sqrt{\frac{{cosec} \theta+\tan \theta}{\sec \theta-\cot \theta}}$
Answer & explanation
Correct answer: option 1
We are given that :-
7 cos²θ + 5 sin²θ - 6 = 0
2 cos²θ + 5 cos²θ + 5 sin²θ = 6
{ using , sin²θ + cos²θ = 1 }
2 cos²θ + 5 = 6
2 cos²θ = 1
cosθ = \(\frac{1}{ √2 }\)
{ cos45º = \(\frac{1}{ √2 }\) }
So, θ = 45º
Now,
\(\sqrt {(cosecθ + tanθ )/ ( secθ - cotθ ) }\)
= \(\sqrt {(cosec 45º + tan 45º )/ ( sec 45º - cot 45º ) }\)
= \(\sqrt {( √2 + 1)/ ( √2 - 1 ) }\)
= \(\sqrt {( √2 + 1)/ ( √2 - 1 ) }\) × \(\sqrt {( √2 + 1)/ ( √2+ 1 ) }\)
= √2 + 1