Question

if x=a secAcosB, y=b secA sinB, z=c tanA prove that x2/a2 + y2/b2 -z2/c2=1

Answer 1

Solution :

$\blue{\sf{LHS= \dfrac{x^{2} }{ {a}^{2} }} + \dfrac{ {y}^{2} }{ {b}^{2} } + \dfrac{ {z}^{2} }{ {c}^{2} }} \\ \\ \implies \sf{ \frac{ (a \: sec \: cos \: B)^{2} }{ {a}^{2} } } + { \frac{ (b \: sec \: cos \: B)^{2} }{ {b}^{2} } } - { \frac{ (c \: tan\:A)^{2} }{ {c}^{2} } } \\ \\ \implies \sf{{ \frac{ a^{2} \: sec^{2} \: cos^{2} \: B }{ {a}^{2} } }} + { \frac{ b^{2} \: sec^{2} \: cos^{2} \: B }{ {b}^{2} } } - { \frac{ c^{2} \: tan^{2} A }{ {c}^{2} } } \\ \\ \implies \sf{sec^{2} A \: cos^{2} B + \: sec^{2} A \: sin ^{2} \: B - tan^{2} \: A}\\ \\ \implies \sf{ sec^{2} \: A \: (cos^{2} \: B + sin^{2} B) - tan^{2} A} \\ \\ \implies \sf{sec^{2} \: A - tan^{2} A \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: (\because{cos B + sin B = 1)}} \\ \\ \implies {\boxed{\sf {\red{1 = RHS}}}}{\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \sf (\because{sec^{2} a - tan^{2} a = 1)} }$

Answer 2

Given:

$\sf \Rrightarrow x = a \sec A \cos B \\ \sf \Rrightarrow y = b \sec A \sin B \\ \sf \Rrightarrow z = c \tan A$

Prove That:

$\sf \Rrightarrow \dfrac{ {x}^{2} }{ {a}^{2} } + \dfrac{ {y}^{2} }{ {b}^{2} } - \dfrac{ {z}^{2} }{ {c}^{2} } = 1$

Proof:

$\sf \to x = a \sec A \cos B$

$\sf \implies \dfrac{x}{a} = \sec A \cos B$

____________

$\sf \to y = b \sec A \sin B$

$\sf \implies \dfrac{y}{b} = \sec A \sin B$

_____________

$\sf \to z = c \tan A$

$\sf \implies \dfrac{z}{c} = \tan A$

_____________

Now,

$\sf \to \dfrac{ {x}^{2} }{ {a}^{2} } + \dfrac{ {y}^{2} }{ {b}^{2} } - \dfrac{ {z}^{2} }{ {c}^{2} } = 1$

$\sf L.H.S = \dfrac{ {x}^{2} }{ {a}^{2} } + \dfrac{ {y}^{2} }{ {b}^{2} } - \dfrac{ {z}^{2} }{ {c}^{2} }$

$\sf = {( \sec A \cos B) }^{2} + {( \sec A \sin B) }^{2} - { \tan}^{2} A$

$\sf = {\sec}^{2} A{\cos }^{2} B + {\sec}^{2} A{ \sin}^{2} B - { \tan}^{2} A$

$\sf = {\sec}^{2} A({\cos }^{2} B + { \sin}^{2} B) - { \tan}^{2} A$

$\sf = {\sec}^{2} A \times 1 - { \tan}^{2} A$

$\sf \begin{cases} \because { \sin}^{2} \theta + { \cos}^{2} \theta = 1 \\ \sf { \sec}^{2} \theta - { \tan}^{2} \theta = 1 \end{cases} \Bigg \}$

$\sf = {\sec}^{2} A - { \tan}^{2} A = 1 = R.H.S$

Hence, Proved

L.H.S = R.H.S

____________

$\sf \looparrowright \dfrac{ {x}^{2} }{ {a}^{2} } + \dfrac{ {y}^{2} }{ {b}^{2} } - \dfrac{ {z}^{2} }{ {c}^{2} } = 1$