Question

cos A / 1 ±sinA = tan (45°+ A/2)
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Answer 1

Answer:

Step-by-step explanation:

CosA/1 ±SinA = Tan(45° ± A/2)

We know that CosA = 1 - Tan²A/2 / 1 + Tan²A/2

                       SinA   = 2TanA/2 / 1 + Tan²A/2.

Consider CosA / 1 + SinA

=> [1 - Tan²A/2 / 1 + Tan²A/2 ] / 1+ [2TanA/2 / 1 + Tan²A/2]

=> [1 - Tan²A/2 / 1 + Tan²A/2 ] / [(1 + Tan²A/2 + 2TanA/2) / 1 + Tan²A/2]

=> 1 - Tan²A/2 / (1 + Tan²A/2 + 2TanA/2) ( ∵ 1 +  Tan²A/2 will cancel out)

//The denominator, it is of form a² + b² + 2ab = (a + b)².

=> 1 - Tan²A/2 / (1 + TanA/2)²

//The numerator is of form a² - b² = (a + b)(a - b)

=> (1 - TanA/2)(1 + TanA/2) / (1 + TanA/2) ( 1 + TanA/2)

//1 + TanA/2 in numerator and denominator will cancel out

=> 1 - TanA/2 / 1 + TanA/2

=> Tan45° - TanA/2 /  1 + Tan45°TanA/2 ( ∵ Tan45° = 1 )

//The above is of form TanA - TanB/ 1 + TanATanB = Tan (A - B)

=> Tan (45° - A/2).

Now Consider CosA / 1 - SinA

=> [1 - Tan²A/2 / 1 + Tan²A/2 ] / 1 - [2TanA/2 / 1 + Tan²A/2]

=> [1 - Tan²A/2 / 1 + Tan²A/2 ] / [(1 + Tan²A/2  - 2TanA/2) / 1 + Tan²A/2]

=> 1 - Tan²A/2 / (1 + Tan²A/2 - 2TanA/2) ( ∵ 1 + Tan²A/2 will cancel out)

//The denominator, it is of form a² + b² - 2ab = (a - b)².

=> 1 - Tan²A/2 / (1 - TanA/2)²

//The numerator is of form a² - b² = (a + b)(a - b)

=> (1 - TanA/2)(1 + TanA/2) / (1 - TanA/2) ( 1 - TanA/2)

//1 - TanA/2 in numerator and denominator will cancel out

=> 1 + TanA/2 / 1 - TanA/2

=> Tan45° + TanA/2 /  1 - Tan45°TanA/2 ( ∵ Tan45° = 1 )

//The above is of form TanA + TanB/ 1 - TanATanB = Tan (A + B)

=> Tan (45° + A/2).

Hence we can prove that CosA/1 ± SinA = Tan(45° ± A/2)