If sec θ = $\frac{65}{63}$ and θ is an acute angle, then the value of 8(cosecθ - cot θ) is :
Answer & explanation
Correct answer: option 1
sec θ = \(\frac{65}{63}\)
{ we know, sec θ = \(\frac{H}{B}\) ]
By using pythagoras theorem,
P² + B² = H²
P² + 63² = 65²
P = 16
Now,
8(cosecθ - cot θ)
= 8 ( \(\frac{65}{16}\) - \(\frac{63}{16}\) )
= 8 ( \(\frac{2}{16}\) )
= 1