Question

how to get answer!??????​

Answer 1

Answer:

$( 1 + \frac{ {z}^{2} }{ {c}^{2} } )$

OPTION A IS CORRECT.

Step-by-step explanation:

$\frac{x}{a} = \sec( \alpha ) \cos( \beta ) \\ \\ \frac{y}{b} = \sec( \alpha ) \sin( \beta ) \\ \\ \frac{ {x}^{2} }{ {a}^{2} } + \frac{ {y}^{2} }{ {b}^{2} } = { \sec( \alpha ) }^{2} ( { \sin( \beta ) }^{2} + { \cos( \beta )) }^{2} \\ \\ \frac{ {x}^{2} }{ {a}^{2} } + \frac{ {y}^{2} }{ {b}^{2} } = { \sec( \alpha ) }^{2} \\ \\ = 1 + { \tan( \alpha) }^{2} \\ \\ =( 1 + \frac{ {z}^{2} }{ {c}^{2} } ) \\ \\ then \: \: \: \: \: \frac{ {x}^{2} }{ {a}^{2} } + \frac{ {y}^{2} }{ {b}^{2} } =( 1 + \frac{ {z}^{2} }{ {c}^{2} } ).$

Answer 2

Step-by-step explanation:

Solution :-

We have,

$\text{\( \rm \sec \theta \cos \phi=\dfrac{x}{a} \) \( \ldots \) (i)}$

$\text{ \( \rm \sec \theta \sin \phi=\dfrac{y}{b} \ldots \) (ii)}$

and

$\text{\( \rm \tan \theta=\dfrac{z}{c} \) ... (iii)}$

Squaring and adding (i) and (ii), we get

$\\ \rm \frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=\sec ^{2} \theta \cos ^{2} \phi+\sec ^{2} \theta \sin ^{2} \phi$

$\rm =\sec ^{2} \theta\left(\cos ^{2} \phi+\sin ^{2} \phi\right)=\sec ^{2} \theta \quad\left[\because \cos ^{2} A+\sin ^{2} A=1\right]$

$\text{ \( \displaystyle \rm =\left(1+\tan ^{2} \theta \right )=\left(1+\frac{z^{2}}{c^{2}}\right) \quad \) [From (iii)]}$

$\\ \rm \therefore \quad\left(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}\right)=\left(1+\frac{z^{2}}{c^{2}}\right)$

$\\ \boxed{\color{violet} \rm \therefore\left(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}\right)=\left(1+\frac{z^{2}}{c^{2}}\right) }$

Therefore Correct option is (A).