Question

Prove that a
COSA
+
1t5inA
1 tsina
1
2 Seca
COSA​

Answer 1

Step-by-step explanation:

Let us start from the LHS

Let us start from the LHS1-sinA/1+sinA

Let us start from the LHS1-sinA/1+sinAOn rationalising the denominator we get

Let us start from the LHS1-sinA/1+sinAOn rationalising the denominator we get1-sinA (1-SinA) /1+sinA(1-SinA)

Let us start from the LHS1-sinA/1+sinAOn rationalising the denominator we get1-sinA (1-SinA) /1+sinA(1-SinA)=(1-SinA)²/ 1² -(Sin²A)

Let us start from the LHS1-sinA/1+sinAOn rationalising the denominator we get1-sinA (1-SinA) /1+sinA(1-SinA)=(1-SinA)²/ 1² -(Sin²A)=(1-SinA)²/ 1 -(Sin²A)

Let us start from the LHS1-sinA/1+sinAOn rationalising the denominator we get1-sinA (1-SinA) /1+sinA(1-SinA)=(1-SinA)²/ 1² -(Sin²A)=(1-SinA)²/ 1 -(Sin²A)=(1-SinA)² /Cos² A

Let us start from the LHS1-sinA/1+sinAOn rationalising the denominator we get1-sinA (1-SinA) /1+sinA(1-SinA)=(1-SinA)²/ 1² -(Sin²A)=(1-SinA)²/ 1 -(Sin²A)=(1-SinA)² /Cos² A=[ 1 -Sin²A = cos²A]

Let us start from the LHS1-sinA/1+sinAOn rationalising the denominator we get1-sinA (1-SinA) /1+sinA(1-SinA)=(1-SinA)²/ 1² -(Sin²A)=(1-SinA)²/ 1 -(Sin²A)=(1-SinA)² /Cos² A=[ 1 -Sin²A = cos²A]=(1-SinA/CosA)²

Let us start from the LHS1-sinA/1+sinAOn rationalising the denominator we get1-sinA (1-SinA) /1+sinA(1-SinA)=(1-SinA)²/ 1² -(Sin²A)=(1-SinA)²/ 1 -(Sin²A)=(1-SinA)² /Cos² A=[ 1 -Sin²A = cos²A]=(1-SinA/CosA)²=(1/CosA-SinA/CosA)²

Let us start from the LHS1-sinA/1+sinAOn rationalising the denominator we get1-sinA (1-SinA) /1+sinA(1-SinA)=(1-SinA)²/ 1² -(Sin²A)=(1-SinA)²/ 1 -(Sin²A)=(1-SinA)² /Cos² A=[ 1 -Sin²A = cos²A]=(1-SinA/CosA)²=(1/CosA-SinA/CosA)²=(SecA-tanA)²

Let us start from the LHS1-sinA/1+sinAOn rationalising the denominator we get1-sinA (1-SinA) /1+sinA(1-SinA)=(1-SinA)²/ 1² -(Sin²A)=(1-SinA)²/ 1 -(Sin²A)=(1-SinA)² /Cos² A=[ 1 -Sin²A = cos²A]=(1-SinA/CosA)²=(1/CosA-SinA/CosA)²=(SecA-tanA)²= RHS

Let us start from the LHS1-sinA/1+sinAOn rationalising the denominator we get1-sinA (1-SinA) /1+sinA(1-SinA)=(1-SinA)²/ 1² -(Sin²A)=(1-SinA)²/ 1 -(Sin²A)=(1-SinA)² /Cos² A=[ 1 -Sin²A = cos²A]=(1-SinA/CosA)²=(1/CosA-SinA/CosA)²=(SecA-tanA)²= RHSHence proved

Answer 2

Answer:

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