If $2cos^2 \theta = 3(1 - sin\theta ), 0° < \theta < 90°, $ then what is the value of $(tan2\theta +cosec3\theta -sec2\theta )$ ?
Answer & explanation
Correct answer: option 2
2 cos²θ = 3 ( 1 - sinθ )
{ sin²θ + cos²θ= 1 }
2 (1-sin²θ) = 3 ( 1 - sinθ )
2 - 2sin²θ = 3 - 3sinθ
2sin²θ - 3sinθ + 1 = 0
2sin²θ - 2sinθ - sinθ + 1 = 0
2sinθ( sinθ - 1 ) - 1 ( sinθ - 1 ) = 0
( sinθ - 1 ).( 2sinθ - 1 ) = 0
Either sinθ - 1 = 0 OR 2sinθ - 1 = 0
sinθ = 1 ( not possible because 0 < θ < 90º )
So, 2sinθ - 1 = 0
sinθ = \(\frac{1}{2}\)
{ sin60º = \(\frac{1}{2}\) }
θ = 30º
Now,
( tan 2θ + cosec 3θ - sec 2θ )
= tan 60º + cosec 90º - sec 60º
= √3 + 1 - 2
= √3 - 1