Simplify: $\frac{(\sin \theta+\sec \theta)^2+(\cos \theta+{cosec} \theta)^2}{(1+\sec \theta {cosec} \theta)^2}, 0^{\circ}<\theta<90^{\circ}$
Answer & explanation
Correct answer: option 2
\(\frac{(sinθ + secθ)² + (cosθ + cosecθ)²}{(1 +secθ. cosecθ)²}\)
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