Trigonometry

Question

proove that sec sq theeta = cosec sq theeta > = 4

Answer 1

I think this is meant to read: prove sec²θ+cosec²θ≥4.

1/cos²θ+1/sin²θ=(sin²θ+cos²θ)/(sinθcosθ)²=

1/(sinθcosθ)².

sin(2θ)=2sinθcosθ, so sinθcosθ=½sin(2θ) and (sinθcosθ)²=½²sin²(2θ)=¼sin²(2θ).

The highest magnitude |sin(2θ)|=1 when 2θ=90° or 270°, for example. So maximum sin²(2θ)=1, so maximum (sinθcosθ)²=¼. The reciprocal of this therefore has a minimum value of 1/¼=4, therefore:

1/(sinθcosθ)²≥4, and sec²θ+cosec²θ≥4 QED