If $cos \theta = \frac{p^2-1}{p^2+1}, 0° < θ < 90°,$ then cosecθ is equal to :
Answer & explanation
Correct answer: option 3
cos θ = \(\frac{p² - 1}{p² + 1}\)
{ cos A = \(\frac{B}{H}\) }
B = p² - 1 & H = p² + 1
By using pythagoras theorem,
P² + B² = H²
P² + ( p² - 1 )² = ( p² + 1 )²
P = \(\sqrt { ( p² + 1 )² -( p² - 1 )² }\)
P = \(\sqrt { 4p² }\)
P = 2p
Now,
cosec θ = \(\frac{H}{P}\)
= \(\frac{ p² + 1}{2p}\)