Question

3 cot theta equals to 4show that 1 minus 10 squared theta divided by 1 + 10 square theta equals to cos square theta minus sin square theta​

Answer 1

Answer:

the anser is 4 by 5

Step-by-step explanation:

the explanation is given in the picture

uploaded above

Answer 2

Answer with Step-by-step explanation:

$3cot\theta=4$

$cot\theta=\frac{4}{3}$

We know that

$tan\theta=\frac{1}{cot\theta}$

Using the formula

$tan\theta=\frac{1}{\frac{4}{3}}=\frac{3}{4}$

$1+tan^2\theta=sec^2\theta$

Substitute the values

$1+(\frac{3}{4})^2=sec^2\theta$

$1+\frac{9}{16}=sec^2\theta$

$\frac{16+9}{16}=sec^2\theta$

$sec\theta=\sqrt{\frac{25}{16}}=\frac{5}{4}$

$cos\theta=\frac{1}{sec\theta}=\frac{1}{\frac{5}{4}}=\frac{4}{5}$

$sin\theta=\sqrt{1-cos^2\theta}=\sqrt{1-(\frac{4}{5})^2}$

$sin\theta=\sqrt{1-\frac{16}{25}}=\sqrt{\frac{25-16}{25}}=\sqrt{\frac{9}{25}}=\frac{3}{5}$

LHS:

$\frac{1-tan^2\theta}{1+tan^2\theta}=\frac{1-(\frac{3}{4})^2}{1+(\frac{3}{4})^2}$

$\frac{1-tan^2\theta}{1+tan^2\theta}=\frac{1-\frac{9}{16}}{1+\frac{9}{16}}$

$\frac{1-tan^2\theta}{1+tan^2\theta}=\frac{\frac{16-9}{16}}{\frac{16+9}{16}}$

$\frac{1-tan^2\theta}{1+tan^2\theta}=\frac{7}{25}$

RHS:

$cos^2\theta-sin^2\theta=(\frac{4}{5})^2-(\frac{3}{5})^2$

$cos^2\theta-sin^2\theta=\frac{16}{25}-\frac{9}{25}=\frac{16-9}{25}=\frac{7}{25}$

LHS=RHS

Hence, proved.

#Learn more:

https://brainly.in/question/752610:Answered by Aroy