Question

any teacher give me the answer

Answer 1

Answer:

$\frac{1-2sin^2\theta}{sin^2\theta(1-sin^2\theta)}$

Step-by-step explanation:

Identities needed:

$cos^2\theta + sin^2\theta = 1$

$\frac{sin^n\theta}{cos^n\theta}=tan^n\theta$

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

Convert into one fraction, using our third identity.

$cot^2\theta-tan^2\theta$

$\frac{1}{tan^2\theta} - tan^2\theta$

Multiply the right hand term by $\frac{tan^2\theta}{tan^2\theta}$ so we can simplify.

$\frac{1-tan^4\theta}{tan^2\theta}$

Now, let's write our tans in terms of sin and cos, using our second identity.

$\frac{1-\frac{sin^4\theta}{cos^4\theta}}{\frac{sin^2\theta}{cos^2\theta}}$

And simplify using division rules, and then simplify by expanding the numerator:

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

$\frac{cos^2\theta-\frac{sin^4\theta}{cos^2\theta}}{sin^2\theta}$

Now we can substitute cos for sin, using our first identity.

$\frac{1-sin^2\theta-\frac{sin^4\theta}{1-sin^2\theta}}{sin^2\theta}$

Lets divide by $sin^2\theta$

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

And convert it back into one fraction:

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

$\frac{(1-sin^2\theta)^2-sin^4\theta}{sin^2\theta-sin^4\theta}$

Expand the binomial:

$\frac{1-2sin^2\theta+sin^4\theta - sin^4\theta}{sin^2\theta(1-sin^2\theta)}$

$\frac{1-2sin^2\theta}{sin^2\theta(1-sin^2\theta)}$