Question

(1+tana)2+(1-tana)2=2sec2a

Answer 1

Answer:

2 sec^2 A

Step-by-step explanation:

    Using 1+ tan^2 A = sec^2 A

⇒ ( 1 + tanA )^2 + ( 1 - tanA )^2

⇒ [ ( 1 )^2 + ( tanA )^2 + 2( tanA )( 1 ) ] + [ ( 1 )^2 + ( tanA )^2 - 2( tanA )( 1 ) ]

⇒ [ 1 + tan^2 A + 2tanA ] + [ 1 + tan^2 A - 2tanA ]

⇒ 1 + tan^2 A + 2tanA + 1 + tan^2A - 2tanA

⇒ 1 + 1 + tan^2 A + tan^2 A

⇒ 2( 1 + tan^2 A )

⇒ 2( sec^2 A )

⇒ 2sec^2 A    

       Proved.

Answer 2

Answer:

$\huge\boxed{\fcolorbox{blue}{orange}{HELLO\:MATE}}$

Step-by-step explanation:

GIVEN:

$(1+tanA)^{2} + (1-tanA)^{2}=2sec^{2}A$

PROOF:

$(1+tanA)^{2} + (1-tanA)^{2}=2sec^{2}A$

LHS:

=$(1+tanA)^{2} + (1-tanA)^{2}$

=$1 + tan^{2}A +2tanA +1+tan^{2}A -2tanA$

$\large\green{\boxed{ (a+b) ^{2}=a^{2}+b^{2}+2ab}}$

$\large\green{\boxed{ (a-b) ^{2}=a^{2}+b^{2}-2ab}}$

=$1 + tan^{2}A +1+tan^{2}A$

=$2 + 2tan^{2}A$

=$2(1+tan^{2}A)$

$\large\red{\boxed{ sec^{2}A=1+tan^{2}A}}$

On substituting this value in above expression,we have:

=$2sec^{2}A$

=LHS

$\huge\purple{\boxed{HENCE\: LHS= RHS}}$