Question

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❶ If m² + m'² + 2mm' cosθ = 1, n² + n'² + 2nn' cosθ = 1 and mn + m'n' + (mn' + m'n) cosθ = 0, prove that m² + n² = cosec²θ

❷ If a cosθ – b sin θ = c, show that a sinθ + b cosθ = ± $\sf{\sqrt{a^2 + b^2 - c^2}}$

Answer with proper explanation.
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Answer 1

Solution (1)

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Given :

• m² + m'² + 2 mm' cos θ = 1
• n² + n'² + 2 nn' cos θ = 1
• mn + m'n' + ( mn' + m'n ) cos θ = 0  ____equation ①

To prove :

• m² + n² = cosec²θ

Proof :

As given

→ m² + m'² + 2 mm' cos θ = 1

adding and subtracting m²cos²θ in LHS

→ m² + m'² + 2 mm' cos θ + m²cos²θ - m²cos²θ  = 1

Rearranging the terms

→ ( m'² + m²cos²θ + 2 mm' cos θ ) + m² - m²cos²θ = 1

using algebraic identity [ (a + b)² = a² + b² + 2 ab ]

→ ( m' + m cos θ )² + m² ( 1 - cos²θ ) = 1

using trigonometric identity [ 1 - cos²A = sin²A ]

→ ( m' + m cos θ )² + m² sin²θ = 1

→ ( m' + m cos θ )² = 1 - m² sin² θ  ___equation ②

Similarly in n² + n'² + 2 nn' cos θ = 1

by adding and subtracting n²cos²θ in LHS and solving in the same way

we will get,

→  ( n' + n cos θ )² = 1 - n² sin² θ  ___equation ③

Now,

multiplying equation ② and ③

→  ( m' + m cos θ )²  ( n' + n cos θ )² = ( 1 - m² sin² θ ) ( 1 - n² sin² θ )

→  [( m' + m cos θ ) ( n' + n cos θ )]² =  ( 1 - m² sin² θ ) ( 1 - n² sin² θ )

→ ( m'n' + m'n cos θ + mn' cos θ + mn cos²θ )² = 1 - n² sin² θ -  m² sin² θ + m²n²sin⁴θ

→ [{m'n' + cos θ ( m'n + mn' )}+ mn cos²θ]² = 1 - sin²θ ( m² + n² ) + m²n²sin⁴θ

Using equation ① in LHS as ; m'n' + ( mn' + m'n ) cosθ = - mn

→ [ ( - mn ) + mn cos²θ ]² =  1 - sin²θ ( m² + n² ) + m²n²sin⁴θ

→ [ - mn ( 1 - cos²θ ) ]² =  1 - sin²θ ( m² + n² ) + m²n²sin⁴θ

using trigonometric identity [ 1 - cos²A = sin²A ]

→ ( - mn sin²θ )² =  1 - sin²θ ( m² + n² ) + m²n²sin⁴θ

→ m²n²sin⁴θ = 1 - sin²θ ( m² + n² ) + m²n²sin⁴θ

→ 0 = 1 -  sin²θ ( m² + n² )

→   sin²θ ( m² + n² ) = 1

→ m² + n² = 1/sin²θ

→ m² + n² = cosec²θ

PROVED.

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Solution (2)

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Given :

• a cosθ - b sinθ = c

To prove :

• a sin θ + b cos θ = ± √( a² + b² - c² )

Proof :

As given

→  a cosθ – b sin θ = c

squaring both sides

→ (  a cosθ – b sin θ )² = c²

using algebraic identity [  (a - b)² = a² + b² - 2 ab ]

→ a²cos²θ + b²sin²θ - 2 ab sinθ cosθ = c²

putting [cos²θ = 1 - sin²θ] and [sin²θ = 1 - cos²θ]

→  a²( 1 - sin²θ ) + b²( 1 - cos²θ ) - 2 ab sinθ cosθ = c²

→ a² - a²sin²θ + b² - b²cos²θ - 2 ab sinθ cosθ = c²

→ a² + b² = c² +  a²sin²θ + b²cos²θ + 2 ab sinθ cosθ

→ a² + b² - c² = a²sin²θ + b²cos²θ + 2 ab sinθ cosθ

using algebraic identity [ a² + b² + 2 a b = ( a + b )² ] in RHS

→ a² + b² - c² = ( a sinθ + b cosθ )²

→  ( a sinθ + b cosθ )² =  a² + b² - c²

→ a sinθ + b cosθ = ± √( a² + b² - c²)

PROVED.

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Answer 2

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