Question

If cot theta +tan theta=x and sec theta-cos theta=y, prove that (x2)2/3 - (xy2)2/3 =1

Answer 1

here is your answer hope you understand my writing and mark me as brainliest

Answer 2

Answer: Proved

Step-by-step explanation:  Given that

$\cot \theta=x,~~~\tan \theta=y.$

We are to prove that

$(x^2y)^\frac{2}{3}-(xy^2)^\frac{2}{3}=1.$

We have

$L.H.S.\\\\=(x^2y)^\frac{2}{3}-(xy^2)^\frac{2}{3}\\\\=\{(\cot\theta+\tan\theta)^2(\sec\theta-\cos\theta)\}^\frac{2}{3}-\{(\cot\theta+\tan\theta)(\sec\theta-\cos\theta)^2\}^\frac{2}{3}\\\\=\left(\dfrac{\cos\theta}{\sin\theta}+\dfrac{\sin\theta}{\cos\theta}\right)^\frac{4}{3}\left(\dfrac{1}{\cos\theta}-\cos\theta\right)^\frac{2}{3}-\left(\dfrac{\cos\theta}{\sin\theta}+\dfrac{\sin\theta}{\cos\theta}\right)^\frac{2}{3}\left(\dfrac{1}{\cos\theta}-\cos\theta\right)^\frac{4}{3}\\\\=\left(\dfrac{\cos^2\theta+\sin^2\theta}{\sin\theta\cos\theta}\right)^\frac{4}{3}\left(\dfrac{\sin^2\theta}{\cos\theta}\right)^\frac{2}{3}-\left(\dfrac{\cos^2\theta+\sin^2\theta}{\sin\theta\cos\theta}\right)^\frac{2}{3}\left(\dfrac{\sin^2\theta}{\cos\theta}\right)^\frac{4}{3}\\\\=\left(\dfrac{(\cos^2\theta+\sin^2\theta)^2}{\sin^2\theta\cos^2\theta}\right)^\frac{2}{3}\left(\dfrac{\sin^2\theta}{\cos\theta}\right)^\frac{2}{3}-\left(\dfrac{\cos^2\theta+\sin^2\theta}{\sin\theta\cos\theta}\right)^\frac{2}{3}\left(\dfrac{\sin^4\theta}{\cos^2\theta}\right)^\frac{2}{3}\\\\=\dfrac{1}{\cos^2\theta}-\dfrac{\sin^2\theta}{\cos^2\theta}\\\\=\sec^2\theta-\tan^2\theta\\=1\\\\=R.H.S.$

Hence proved.