Question

find value of tan²/tan²-1 + cosec²/sec²-cosec²
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Answer 1

Answer:

$\frac{1}{sin^2-cos^2}$

Step-by-step explanation:

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

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Answer 2

Answer:

1/(sin²-cos²)