Question

Answer question number 3​

Answer 1

Here is ur ans
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Answer 2

Answer: $\frac{q^2}{p\sqrt{q^2-p^2}}$

Step-by-step explanation:

Given,

$q\cos\theta=p\\\;\\\cos\theta=\frac{p}{q}$

We know  that,

$\sin^2\theta=1-\cos^2\theta\\\;\\\sin^2\theta=1-(\frac{p}{q})^2\\\;\\\sin^2\theta=1-\frac{p^2}{q^2}\\\;\\\sin^2\theta=\frac{q^2-p^2}{q^2}\\\;\\\sin\theta=\frac{\sqrt{q^2-p^2}}{q}$

Now,

$\sin\theta.\cos\theta=\frac{\sqrt{p^2-q^2}}{q}\times\frac{p}{q}\\\;\\\sin\theta.\cos\theta=\frac{p\sqrt{p^2-q^2}}{q^2}$

And,

$\sin^2\theta-\cos^2\theta=1-\frac{p^2}{q^2}-\frac{p^2}{q^2}\\\;\\\sin^2\theta-\cos^2\theta=1$

We have,

$\tan\theta-\cot\theta\\\;\\=\frac{\sin\theta}{\cos\theta}-\frac{\cos\theta}{\sin\theta}\\\;\\=\frac{\sin^2\theta-\cos^2\theta}{\sin\theta.\cos\theta}\\\;\\=\frac{1}{\frac{p\sqrt{q^2-p^2}}{q^2}}\\\;\\=\frac{q^2}{p\sqrt{q^2-p^2}}$