Question

help please maths questions for 9th​

Answer 1

$\overline{\underline{\boxed{\sf GIVEN \darr}}}$

1) (sin θ + cos θ)² - (sin θ - cos θ)² = 4 sin θ cos θ

2) (sin⁴ θ - cos⁴ θ)/(sin² θ - cos² θ) = 1

$\overline{\underline{\boxed{\sf SOLUTION \darr}}}$

1) By using square relations we get

sin θ + cos θ = 1

So, keeping this in mind we get the value of this problem

• (sin θ + cos θ)² - (sin θ - cos θ)² = 4 sin θ cos θ

By simplifying we get

• sin² θ + 2 sin θ cos θ - (sin² θ - 2 sin θ cos θ + cos² θ = 4 sin θ cos θ

• sin² θ + 2 sin θ cos θ - sin² θ - 2 sin θ cos θ + cos² θ = 4 sin θ cos θ

• 2 sin θ cos θ + 2 sin θ cos θ = 4 sin θ cos θ

• 4 sin θ cos θ = 4 sin θ cos θ

• LHS = RHS

Hence, Proved !!

2) (sin⁴ θ - cos⁴ θ)/(sin² θ - cos² θ) = 1

By cross multiplying we get

• (sin⁴ θ - cos⁴ θ) = 1 × (sin² θ - cos² θ)

• (sin⁴ θ - cos⁴ θ) = (sin² θ - cos² θ)

sin⁴ θ - cos⁴ θ can be written as (sin² θ)² - (cos² θ)²

• (sin² θ)² - (cos² θ)² = sin² θ - cos² θ

Now applying a² - b² = (a - b)(a + b)

• (sin² θ - cos² θ)(sin² θ + cos² θ) = sin² θ - cos² θ

Now transposing sin² θ - cos² θ ,

• sin² θ + cos² θ = (sin² θ - cos² θ)/(sin² θ - cos² θ)

(sin² θ - cos² θ)/(sin² θ - cos² θ) = 1

• sin² θ + cos² θ = 1

By using square relations :

sin² θ + cos² θ is

So,

• 1 = 1

• LHS = RHS

Hence, Proved !!

$\overline{ \underline{ \boxed{ \sf LEARN \: MORE \: \darr}}}$

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$\star \: \underline{ \boxed{\purple{ \mathfrak{ {a}^{2} + {b}^{2} = {(a + b)}^{2} - 2ab }}}}$

$\star \: \underline{ \boxed{ \orange{\mathfrak{ {a}^{2} + {b}^{2} = {(a - b)}^{2} + 2ab}}}}$

$\star \: \underline{ \boxed{ \pink{\mathfrak{ {(a + b)}^{2} + {(a - b)}^{2} = 2( {a}^{2} + {b}^{2} )}}}}$

$\star \: \underline{ \boxed{ \gray{\mathfrak{ {(a + b)}^{2} - {(a - b)}^{2} = 4ab }}}}$

$\star \: \underline{ \boxed{ \green{\mathfrak{ {(a + b + c)}^{2} = {a}^{2} + {b}^{2} + {c}^{2} + 2(ab + bc + ca) }}}}$

$\star \: \underline{ \boxed{ \mathfrak{ \blue{{(a + b - c)}^{2} = {a}^{2} + {b}^{2} + {c}^{2} + 2(ab - bc - ca) }}}}$