Question

prove that sin square theta divided by cos square theta + cos square theta divided by sin square theta is equal to secant squared theta minus cosecant squared theta minus 2

Answer 1

we have to prove that ,

$\frac{sin^2\theta}{cos^2\theta}+\frac{cos^2\theta}{sin^2\theta}=sec^2\theta cosec^2\theta-2$

$LHS=\frac{sin^2\theta}{cos^2\theta}+\frac{cos^2\theta}{sin^2\theta}\\\\=\frac{sin^4\theta+cos^4\theta}{sin^2\theta cos^2\theta}\\\\=\frac{(sin^2\theta+cos^2\theta)^2-2sin^2\theta cos^2\theta}{sin^2\theta cos^2\theta}\\\\\textbf{we know},sin^2x+cos^2x=1,\textbf{so},sin^2\theta+cos^2\theta=1\\\\=\frac{1-2sin^2\theta cos^2\theta}{sin^2\theta cos^2\theta}\\\\=\frac{1}{sin^2\theta cos^2\theta}-2\\\\=cosec^2\theta sec^2\theta-2=RHS$

Answer 2

Answer:

Sin^2x/cos^2x +cos^2÷sin^2 = sec^2-cosec^2-2

Let LHS be sin^2÷cos^2+cos^2÷sin ^2 = tan^2+cot^2 =( sec^2-1) +(cosec^2-1) =sec^2-1+cosec^2-1=sec^2+cosec^2-1-1= sec^2+cosec^2 -2hence proved it's not possible to prove sec^2-cosec^2-2