Question

prove that 1 + tan squared theta upon 1 + cot squared theta is equals to 1 minus 10 theta upon 1 minus cot theta square is equals to tan square theta

Answer 1

Answer:

To Prove:

$\frac{1+tan^2\,\theta}{1+cot^2\,\theta}=(\frac{1-tan\,\theta}{1-cot\,\theta})^2=tan^2\,\theta$

Consider,

$\frac{1+tan^2\,\theta}{1+cot^2\,\theta}$

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

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

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

$=\frac{sin^2\,\theta}{cos^2\,\theta}$

$=tan^2\,\theta$

Now Consider,

$(\frac{1-tan\,\theta}{1-cot\,\theta})^2$

$=\frac{(1-tan\,\theta)^2}{(1-cot\,\theta)^2}$

$=\frac{1+tan^2\,\theta-2tan\,\theta}{1+cot^2\,\theta-2cot\,\theta}$

$=\frac{sec^2\,\theta-2tan\,\theta}{cosec^2\,\theta-2cot\,\theta}$

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

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

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

$=\frac{sin^2\,\theta}{cos^2\,\theta}$

$=tan^2\,\theta$

Hence Proved.

Answer 2

Step-by-step explanation:

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