If sec A$ =\frac{\sqrt{11}}{3}$, then the value of $\frac{cosec^2A + tan^2A}{sin^2+cot^2A}$ is :
Answer & explanation
Correct answer: option 3
secA = \(\frac{√11 }{3}\)
{ cosx = \(\frac{B }{H}\) }
P² + B² = H²
P² + 9 = 11
P = √2
Now,
\(\frac{cosec²A + tan²A }{sin²A + cot²A}\)
= \(\frac{ (√11/√2)² +(√2/3)² }{(√2/√11)² + (3/√2)² }\)
= \(\frac{11/2 + 2/9 }{2/11 + 9/2}\)
= \(\frac{(99+4)/18 }{(4+99)22}\)
= \(\frac{22 }{18}\)
= \(\frac{11 }{9}\)