Question

Sin2/cos2 +cos2/sin2=sec2-cosec2-2

Answer 1

Answer:

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

Step-by-step explanation:

Formula used:

$cos^2\theta+sin^2\theta=1$

Now,

$\frac{sin^2\theta}{cos^2\theta}+\frac{cos^2\theta}{sin^2\theta}\\\\=\frac{sin^2\theta}{cos^2\theta}+\frac{cos^2\theta}{sin^2\theta}\\\\=\frac{1-cos^2\theta}{cos^2\theta}+\frac{1-sin^2\theta}{sin^2\theta}\\\\=\frac{1}{cos^2\theta}-\frac{cos^2\theta}{cos^2\theta}+\frac{1}{sin^2\theta}-\frac{sin^2\theta}{sin^2\theta}\\\\=\frac{1}{cos^2\theta}-1+\frac{1}{sin^2\theta}-1\\\\=sec^2\theta+cosec^2\theta-2$