Prove that :- {(1+cosA)^2 +(1-cosA) ^2} ÷ {sin^2A} = 2 (1+2cot^2A)

Step-by-step explanation:

(a + b)² + (a - b)² = 2(a² + b²)

a = 1 ; b = Cos A

(1 + Cos A)² + (1 - Cos A)² = 2(1² + Cos² A)

                                         = 2(1 +Cos² A)   ----------(I)

LHS = { (1 + Cos A)² + (1 - Cos A)²} ÷ { Sin² A}

      = { 2( 1 + Cos² A )} ÷ Sin² A                  {From (I)}

      = { 2 (Sin² A + Cos² A + Cos² A) } ÷ { Sin² A}    {Here, Sin² A + Cos² A = 1 }

     = 2 (Sin² A + 2Cos² A) }  ÷ { Sin² A}    

      = 2( 1 + 2 Cot²  A) =RHS

x = 17

step-by-step explanation:

in the given figure, two triangles are inscribed in a circle where ∠bae ≅ ∠dae which makes the side be and de equal so we can write it as:

3x - 24 = x + 10

solving this equation for x to get:

3x - x = 10 + 24

2x = 34

x = 34/2

x = 17

therefore, the value of x = 17.

step-by-step explanation:

3024 rhe base is what up in worry about