Question

Anyone pls help me with this trigonometry Question​

Answer 1

$\textsf{\large{\underline{Solution}:}}$

Given:

$\rm \longrightarrow x = a \sin \alpha \cos \beta$

$\rm \longrightarrow y = b \sin \alpha \sin \beta$

$\rm \longrightarrow z = c\cos \alpha$

Therefore:

$\rm \longrightarrow \dfrac{x}{a} = \sin \alpha \cos \beta$

$\rm \longrightarrow \dfrac{y}{b} = \sin \alpha \sin \beta$

$\rm \longrightarrow \dfrac{z}{c} = \cos \alpha$

Now:

$\rm = \dfrac{ {x}^{2} }{ {a}^{2}} + \dfrac{ {y}^{2} }{ {b}^{2} } + \dfrac{ {z}^{2} }{ {c}^{2} }$

$\rm = { \sin}^{2} \alpha \cos^{2} \beta + { \sin}^{2} \alpha \sin^{2} \beta + { \cos}^{2} \alpha$

$\rm = { \sin}^{2} \alpha (\cos^{2} \beta + \sin^{2} \beta )+ { \cos}^{2} \alpha$

$\rm = { \sin}^{2} \alpha \cdot1+ { \cos}^{2} \alpha$

$\rm = { \sin}^{2} \alpha + { \cos}^{2} \alpha$

$\rm = 1$

Therefore:

$\rm \longrightarrow\dfrac{ {x}^{2} }{ {a}^{2}} + \dfrac{ {y}^{2} }{ {b}^{2} } + \dfrac{ {z}^{2} }{ {c}^{2} } = 1$

Which is our required answer.

$\textsf{\large{\underline{Learn More}:}}$

1. Relationship between sides and T-Ratios.

• sin(x) = Height/Hypotenuse
• cos(x) = Base/Hypotenuse
• tan(x) = Height/Base
• cot(x) = Base/Height
• sec(x) = Hypotenuse/Base
• cosec(x) = Hypotenuse/Height

2. Square formulae.

• sin²x + cos²x = 1
• cosec²x - cot²x = 1
• sec²x - tan²x = 1

3. Reciprocal Relationship.

• sin(x) = 1/cosec(x)
• cos(x) = 1/sec(x)
• tan(x) = 1/cot(x)

4. Cofunction identities.

• sin(90° - x) = cos(x)
• cos(90° - x) = sin(x)
• cosec(90° - x) = sec(x)
• sec(90° - x) = cosec(x)
• tan(90° - x) = cot(x)
• cot(90° - x) = tan(x)

5. Even odd identities.

• sin(-x) = -sin(x)
• cos(-x) = cos(x)
• tan(-x) = - tan(x)