$(\sec \theta+\tan \theta)^2+\frac{1+{cosec} \theta}{1-{cosec} \theta}, 0^{\circ}<\theta<90^{\circ} \text { is: }$
Answer & explanation
Correct answer: option 1
( secθ + tanθ )² + \(\frac{1 + cosecθ}{1 - cosecθ}\)
Let us assume that ,
θ = 45º
= ( sec45º + tan45º )² + \(\frac{1 + cosec45º}{1 - cosec45º}\)
= ( √2 + 1 )² + \(\frac{1 + √2}{1 - √2}\)
= ( √2 + 1 )² + \(\frac{1 + √2}{1 - √2}\) × \(\frac{1 + √2}{1 + √2}\)
= ( √2 + 1 )² + \(\frac{( √2 + 1 )²}{1 - (√2)²}\)
= ( √2 + 1 )² + \(\frac{( √2 + 1 )²}{-1}\)
= ( √2 + 1 )² - ( √2 + 1 )²
= 0