Question

(1+tanA+secA) (1+cotA-cosecA)​

Answer 1

Answer:

Step-by-step explanation:

(1+tanA+secA) (1+cotA-cosecA)​

=> [1 + SinA/CosA + 1/CosA] [ 1 + CosA/SinA - 1/SinA]

=> [(CosA + SinA + 1) / CosA] [ (SinA + CosA - 1)/SinA]

=> 1/SinACosA { [(SinA+CosA) + 1][(SinA+CosA) - 1] }

//the expression in curly brackets is of form (a + b)(a-b) => a² - b²

=> 1/SinACosA [ (SinA+CosA)² - 1²]

//(SinA+CosA)² is of form (a + b)² => a² + b² + 2ab

=> 1/SinACosA [ Sin²A + Cos²A + 2SinACosA - 1 ]

//We know that  Sin²A + Cos²A = 1

=> 1/SinACosA [ 1 + 2SinACosA - 1]

=> 2SinACosA/SinACosA

=> 2

Answer 2

Answer:

$(1 + \tan(a) + \sec(a) ) \\ (1 +\cot(a) - cosec(a)) \\ = ( \frac{ \cos(a) + \sin(a) + 1 }{ \cos(a) } )( \frac{ \sin(a) + \cos(a) - 1 }{ \sin(a) } ) \\ = \frac{( \sin(a) + \cos(a)) {}^{2} - 1 {}^{2} }{ \sin(a) \cos(a) } \\ = \frac{1 + 2 \sin(a) \cos(a) - 1 }{ \sin(a) \cos(a) } \\ = 2$

hope it's helpful