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

2 thoughts on “Prove that a<br />COSA<br />+<br />1t5inA<br />1 tsina<br />1<br />2 Seca<br />COSA​”

1. Answer:

   Hello Vedika where do you live

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2. 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

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