Question

If m sin a+ n cos a = p and m cos a - n sin A=q
prove that m²+ n ²= p ²+ q.²

Answer 1

Answer:

Given:

• m sinA + n cosA = p
• m cosA - n sinA = q

$\mathfrak{\underline{Solution:-}}$

➣Squaring first equation:

→ (m sinA + n cosA)² = p²

→ (m sinA)² + (n cosA)² + 2 (m sinA)(n cosA) = p²

→ m²sin²A + n²cos²A + 2mn sinA cosA = p² - (1)

➣Squaring second equation;

→ (m cosA - n sinA)² = q²

→ (m cosA)² + (n sinA)² - 2 (m cosA) (n sinA) = q²

→ m²cos²A + n²sin²A - 2mn sinA cosA = q² - (2)

Now adding both the equations:

➫ p² + q² = m²sin²A + n²cos²A + 2mn sinA cosA + m²cos²A + n²sin²A - 2mn sinA cosA

➫ p²+ q² = m²sin²A + n²cos²A + m²cos²A + n²sin²A

➫ p² + q² = m² (sin²A + cos²A) + n² (cos²A + sin²A)

We know that , sin²A + cos²A = 1 ,

•°• p² + q² = m² + n²

Additional information:

$\begin{lgathered}\boxed{\begin{array}{l}\sf Fundamental\ trigonometric\ identities: \\ sin^2 A + cos^2 A = 1 \\ 1 + tan^2 A = sec^2 A \\ 1 + cot^2 A = cosec^2 A \\ \end{array}}\end{lgathered}$