If $(cos^2 \theta -1) (2sec^2 \theta) + sec^2 \theta + 2tan^2\theta = 2, 0° < \theta < 90°$, then the value of $\frac{(sec\theta+sin\theta)}{(cosec\theta - cos \theta)}$ will be :
Answer & explanation
Correct answer: option 3
( cos²θ - 1 ) (2sec²θ ) + sec²θ + 2 tan²θ = 2
{using formula, sin²θ + cos²θ = 1 }
( -sin²θ ) (2sec²θ ) + sec²θ + 2 tan²θ = 2
( -2tan²θ ) + sec²θ + 2 tan²θ = 2
{using formula, sec²θ - tan²θ = 1 }
tan²θ = 1
{we know, tan45º = 1}
S0 , θ = 45º
Now,
\(\frac{secθ+ sinθ}{cosecθ - cosθ}\)
= \(\frac{sec45º+ sin45º}{cosec45º - cos45º}\)
= \(\frac{√2+ 1/√2}{√2 - 1/√2}\)
= \(\frac{3}{1}\)
= 3