Question

if x/a cos theta + y/b sin theta = m and x/a sin theta- y/b cos theta=n, prove that x^2/a^2+y^2/b^2=m^2+n^2.​

Answer 1

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(Theta is taken as "A")

Given:-

x/a Cos A + y/b sin A = m

On squaring both sides we get,

→ ( x/a Cos A + y/b sin A)² = m²

Using the formula (a + b)² = a² + b² + 2ab in LHS we get,

→ x²/a² cos² A + y²/b² + 2xy/ab sin A Cos A = m² -- equation (1)

Similarly,

x/a sin A - y/b Cos A = n

→ (x/a sin A - y/b Cos A)² = n²

Using the formula (a - b)² = a² + b² - 2ab in LHS we get,

→ x²/a² sin² A + y²/b² cos² A - 2xy/ab sin A Cos A = n² -- equation (2)

Add equations (1) & (2).

→ x²/a² cos² A + y²/b² sin² A + 2xy/ab sin A Cos A + x²/a² sin² A + y²/b² cos² A - 2xy/ab sin A Cos A = m² + n²

→ (x²/a² cos² A + x²/a² sin² A) + (y²/b² sin² A + y²/b² cos² A) = m² + n²

→ x²/a² ( cos² A + sin² A ) + y²/b² (sin² A + cos² A) = m² + n²

Using the identity cos² A + sin² A = 1 in LHS we get,

→ x²/a² (1) + y²/b² (1) = m² + n²

→ x²/a² + y²/b² = m² + n²

Hence, Proved.✔✔

Answer 2

Step-by-step explanation:

(Theta is taken as "A")

Given:-

x/a Cos A + y/b sin A = m

On squaring both sides we get,

→ ( x/a Cos A + y/b sin A)² = m²

Using the formula (a + b)² = a² + b² + 2ab in LHS we get,

→ x²/a² cos² A + y²/b² + 2xy/ab sin A Cos A = m² -- equation (1)

Similarly,

x/a sin A - y/b Cos A = n

→ (x/a sin A - y/b Cos A)² = n²

Using the formula (a - b)² = a² + b² - 2ab in LHS we get,

→ x²/a² sin² A + y²/b² cos² A - 2xy/ab sin A Cos A = n² -- equation (2)

Add equations (1) & (2).

→ x²/a² cos² A + y²/b² sin² A + 2xy/ab sin A Cos A + x²/a² sin² A + y²/b² cos² A - 2xy/ab sin A Cos A = m² + n²

→ (x²/a² cos² A + x²/a² sin² A) + (y²/b² sin² A + y²/b² cos² A) = m² + n²

→ x²/a² ( cos² A + sin² A ) + y²/b² (sin² A + cos² A) = m² + n²

Using the identity cos² A + sin² A = 1 in LHS we get,

→ x²/a² (1) + y²/b² (1) = m² + n²

→ x²/a² + y²/b² = m² + n²

Hence, Proved.✔✔