If A and B are acute angles and sec A = 3; cot B = 4, then the value of $\frac{cosec^2A +sin^2B}{cot^2+sec^2B}$ is :
Answer & explanation
Correct answer: option 2
With angle A,
secA = 3
P² + B² = H²
P² + 1² = 3²
P² = 8
P = 2√2
And with angle B
cotB = 4
P² + B² = H²
1² + 4² = H²
H² = 17
H = √17
Now,
$\frac{cosec^2A +sin^2B}{cot^2A+sec^2B}$
= $\frac{ (3/2√2)² + (1/√17)²}{(1/2√2)² +(√17/4)²}$
= \(\frac{161/136 }{19/16}\)
= \(\frac{322 }{323}\)