Question

X= b sec3 theta and y=a tan3 theta find (x/b) 2/3 - (y/a) 2/3​

Answer 1

Answer

1

Explanation

Given,

x = b sec³∅ and y = a tan³∅!

→ x/b = sec³∅ and y/a = tan³∅

We have to find

$\sf(\frac{x}{b})^{(\frac{2}{3})} - (\frac{y}{a})^{(\frac{2}{3})}$

Putting the value of x/b as sec³∅ and that of y/a as tan³∅ we get

$\sf sec^3\theta^{\frac{2}{3}} - tan^3\theta^{\frac{2}{3}}$

= sec²∅ - tan²∅

= 1

(Using the identity, 1 + tan²∅ = sec²∅.)

Answer 2

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${\underline{\underline{\mathbb{\blue{Given}}}}}$

$x = b \: { \sec \theta }^{3}$

$y = a \: { \tan \theta }^{3}$

Now finding the value of x/b and y/a

$\frac{x}{b} = { \sec \theta }^{3}$

$\frac{y}{a} = { \tan \theta}^{3}$

We have to find out the value of (x/b)⅔ and (y/a)⅔

Putting the required values here :-

$(\frac{ {x} }{b} )^{ \frac{2}{3} } - (\frac{y}{a} )^{ \frac{2}{3} } = ( { \sec\theta}^{3} )^{ \frac{2}{3} } - {( \tan\theta}^{3} ) ^{ \frac{2}{3} }$

→ sec theta² - tan theta ²

As we know that tan theta² + 1 = sec theta²

So , 1 = sec theta ² - tan theta²

→ 1 is the required value