Question

prove that tan squared theta by tan squared theta minus 1 + cosec square theta by sec square theta minus cosec square theta is equals to 1 by sin square theta minus cos square theta ​

Answer 1

Answer:

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

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

Step-by-step explanation:

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

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

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

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

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

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

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

Therefore.,

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

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

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Answer 2

Step-by-step explanation:

arrow marks indicates how the multiplication done