Question

Sec^4A-tan^4A=1 +2tan^2 A

Answer 1

sec^4A - tan^4A
(sec^2A)^2 - (tan^2A)^2
(sec^2A + tan^2A) (sec^2A - tan^2A)
because (a^2 - b^2) = (a + b) (a - b)
so, (sec^A + tan^2A).1
(1 + tan^2A + tan^2A)
(1 + 2tan^2A).

Answer 2

$\displaystyle \sf{ {sec}^{4}A - {tan}^{4}A = 1 + 2{tan}^{2}A } \: \: is \bf \: proved$

Given :

The expression

$\displaystyle \sf{ {sec}^{4}A - {tan}^{4}A = 1 + 2{tan}^{2}A }$

To find :

Prove the expression

Solution :

Step 1 of 2 :

Write down the given expression

The given expression is

$\displaystyle \sf{ {sec}^{4}A - {tan}^{4}A = 1 + 2{tan}^{2}A }$

Step 2 of 2 :

Prove the expression

LHS

$\displaystyle \sf{ = {sec}^{4}A - {tan}^{4}A }$

$\sf = {({sec}^{2}A )}^{2} -{( {tan}^{2}A )}^{2}$

$\displaystyle \sf = ({sec}^{2}A + {tan}^{2}A ) ({sec}^{2}A -{tan}^{2}A )\: \: \: \bigg[ \: \because \: {a}^{2} - {b}^{2} = (a + b)(a - b)\bigg]$

$\sf = (1 + {tan}^{2}A + {tan}^{2}A ) (1 + {tan}^{2}A-{tan}^{2}A )\: \: \: \bigg[ \: \because \:{sec}^{2}A = 1 + {tan}^{2}A \bigg]$

$\displaystyle \sf{ = (1 +2 {tan}^{2}A ) \times 1 }$

$\displaystyle \sf{ = 1 +2 {tan}^{2}A }$

= RHS

Hence the proof follows

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