Question

eleminate a from the following
x = cot a + tan a;
y = sec a - cos a​

Answer 1

Answer:

$\large{\boxed{\sf{\red{hope \: this \: will \: help \: you}}}} >$

PLEASE FIND THE ATTACHMENT.

Answer 2

Answer:

$\huge\underline{ \underline{ \red {your \: \: \: answer}}}$

Step-by-step explanation:

$\bf\red{ \mapsto \:x = cot \: a + tan \: a } \\ \\ \bf\red{ \mapsto \: x = \frac{cos \: a}{sin \: a} + \frac{sin \: a}{cos \: a} } \\ \\ \bf\red{ \mapsto \:x = \frac{ {cos}^{2}a + {sin}^{2}a }{sin \: a \: cos \: a} } \\ \\ \bf\red{ \mapsto \: x = \frac{1}{sin \: a \: cos \: a} } \\ \\ \bf\purple{ \mapsto \: y = sec \: a - cos \: a} \\ \\ \bf\purple{ \mapsto \:y = \frac{1}{cos \: a} -cos \: a } \\ \\ \bf\purple{ \mapsto \:y = \frac{1 - {cos}^{2}a }{cos \: a} } \\ \\ \bf\purple{ \mapsto \: y = \frac{ {sin}^{2}a }{cos \: a} } \\ \\ \bf\pink{ \mapsto \: {x}^{2}y = \frac{1}{ {sin}^{2}a {cos}^{2}a } \times \frac{ {sin}^{2}a}{cos \: a} = \frac{1}{ {cos}^{3} } } \\ \\ \bf\pink{ \mapsto \: \frac{x}{y} = \frac{1}{sin \: a \: cos \: a} \times \frac{cos \: a}{ {sin}^{2}a } } \\ \\ \bf\pink{ \implies \: \frac{1}{{sin}^{3}a } } \\ \\ \bf\pink{ \mapsto \: {cos}^{3} = {x}^{2}y \implies {cos}^{2}a = ({{x}^{2}y}){^ \frac{1}{3} }} \\ \\ \bf\pink{ \mapsto \: {sin}^{3} a = \frac{y}{x} \implies \: sin \: a = (\frac{y}{x})^{ \frac{1}{3} } } \\ \\ \bf\pink{ \mapsto \: {sin}^{2} a + { cos}^{2} a = 1} \\ \\ \bf\pink{ \mapsto \:( {x}^{2}y) ^{ \frac{2}{3}} +(\frac{y}{x} )^{ \frac{2}{3} } = 1 } \\ \\ \large\boxed{ \fcolorbox{red}{pink}{hope \: it \: help \: you}}$