Question

prove that
tan2 theta/tan 2 theta-1 + cosec 2 theta/sec 2 theta - cosec 2 theta =1/sin 2 theta - cos2 theta

Answer 1

Answer:

I have attached the picture for better understanding:

Answer 2

Answer:

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

Step-by-step explanation:

$LHS=\frac{tan^{2}\theta}{tan^{2}\theta-1}+\frac{cosec^{2}\theta}{sec^{2}\theta-cosec^{2}\theta}$

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

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

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

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

$=\frac{1}{sin^{2}\theta-cos^{2}\theta}\\=RHS$

Therefore,

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

•••♪