Simplify: $\frac{(\sin \theta+\sec \theta)^2+(\cos \theta+{cosec} \theta)^2}{(1+\sec \theta {cosec} \theta)^2}, 0^{\circ}<\theta<90^{\circ}$

1

\(\frac{(sinθ + secθ)² + (cosθ + cosecθ)²}{(1 +secθ. cosecθ)²}\)

Let us assume that ,

θ= 45º

= \(\frac{(sin45º + cos45º)² + (cos45º + cosec45º)²}{(1 +sec45º. cosec45º)²}\)

= \(\frac{(1/√2 + √2)² + (1/√2 + √2)²}{(1 +√2. √2)²}\)

= \(\frac{(3/√2)² + (3/√2)²}{(1 +2)²}\)

= \(\frac{9/2 + 9/2}{9}\)

= \(\frac{9}{9}\)

= 1