Respuesta :

Adithya027

Answer:

To Prove,

(1+tanA-secA)×(1+tanA+secA)=2tanA

LHS,

(1+tanA-secA)×(1+tanA+secA)

Let's Expand the brackets,

=1+tanA-secA+tanA+tan²A+tanA×(-secA)+secA+(tanA×secA)-sec²A

=1+tanA+tanA-secA+secA-tanAsecA+tanAsecA+tan²A-sec²A

=1+2tanA+tan²A-sec²A [secA-secA=0 and tanAsecA-tanAsecA=0]

=1+2tanA-1 [sec²A-tan²A=1; so, -sec²A+tan²A=-1]

=2tanA

=RHS

∴LHS=RHS

∴Proven

Answer:

see explanation

Step-by-step explanation:

using the trigonometric identity

• 1 + tan²x = sec²x

Consider the left side

(1 + tanA - secA)(1 + tanA + secA) ← expand using FOIL

= (1 + tanA)² + secA(1 + tanA) - secA(1 + tanA) - sec²A ← collect like terms

= 1 + 2tanA + tan²A - sec²A

= 1 + tan²A - sec²A + 2tanA

= sec²A - sec²A + 2tanA

= 2tanA

= right side ⇒ proven