Question

Given that:-
• x Cos α- y Sin α=p
• x Sin α +y Cos α=q

Find the value of x²+y².

Class-10th
Trigonometry


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Answer 1

★ ANSWER:-

⇒ P²+Q²  

 ★ SOLUTION:-

 Given equations:-

• x cos α- y sin α=p

• x sin α + y cos α=q  

By squaring these equations and adding both we get:

 (x cos α - y sin α)²+(x sin α + y cos α)² =p²+q²  

 By applying Algebric identities we get:

 ⇒ (X cos α)²+(y sin α)²-2(x cos α)(y sin α)+(x sin α)²+(y cos α)²+2(x sin α)(y cos α)=p²+q²

  ⇒ x²Cos²α+y²sin²α-2 xy cos α sin α+x²sin²α+y²cos²α+2xy sin α cos α=p²+q².

   Here -2xy sin α cos α and + 2 xy sin α cos α will be cancelled.

   ⇒ x²(cos²α+sin²α)+y²(sin²α+cos²α)=p²+q²  

 ⇒ x²(1)+ y²(1)=p²+ q²  

 Because sin²a+cos²a=1  [identity]

   ⇒ x²+y²=p²+q²  So the answer is p²+q².

Answer 2

Required Answer:-

Given:

• x cos α - y sin α = p
• x sin α + y cos α = q

To Find:

• The value of x² + y²

Formula Used:

• sin²α + cos²α = 1

Solution:

Given that,

→ x cos α - y sin α = p — (i)

→ x sin α + y cos α = q — (ii)

Squaring both sides of equation (i), we get,

→ x²cos²α + y²sin² α - 2xy sin α cos α = p² — (iii)

Squaring both sides of equation (ii), we get,

→ x²sin²α + y²cos²α + 2xy sin α cos α = q² — (iv)

Adding (iii) and (iv), we get,

→ x²sin²α + x²cos²α + y²sin²α + y²cos²α = p² + q²

Taking x² and y² as common, we get,

→ x²(sin²α + cos²α) + y²(sin²α + cos²α) = p² + q²

Now, we know that,

→ sin²α + cos²α = 1

→ x² × 1 + y² × 1 = p² + q²

→ x² + y² = p² + q²

★ Therefore, x² + y² = p² + q²

Answer:

• x² + y² = p² + q².