Question

sec⁴ -tan ⁴ = 1 + 2 tan², prove the equation ​

Answer 1

Answer:

Step-by-step explanation:

We have,

$\tt{sec^4(\theta)-tan^4(\theta)}$

$\tt{=\left\{sec^2(\theta)-tan^2(\theta)\right\}\left\{sec^2(\theta)+tan^2(\theta)\right\}}$

$\tt{=1\cdot\left\{1+tan^2(\theta)+tan^2(\theta)\right\}}$

$\tt{=1+2\,tan^2(\theta)}$

Answer 2

Given Question

Prove that

$\rm :\longmapsto\: {sec}^{4}x - {tan}^{4}x = 1 + 2 {tan}^{2}x$

$\large\underline{\sf{Solution-}}$

Consider LHS

$\rm :\longmapsto\: {sec}^{4}x - {tan}^{4}x$

can be rewritten as

$\rm \:  =  \: {( {sec}^{2}x) }^{2} - {( {tan}^{2} x)}^{2}$

We know,

$\purple{\rm :\longmapsto\:\boxed{\tt{ {x}^{2} - {y}^{2} = (x + y)(x - y) \: }}}$

So, using this identity, we get

$\rm \:  =  \: ( {sec}^{2}x - {tan}^{2}x)( {sec}^{2}x + {tan}^{2}x)$

We know

$\purple{\rm :\longmapsto\:\boxed{\tt{ {sec}^{2}x - {tan}^{2}x = 1}}}$

So, using this identity, we get

$\rm \:  =  \: 1 \times (1 + {tan}^{2}x + {tan}^{2}x)$

$\rm \:  =  \: 1 + {2tan}^{2}x$

Hence,

$\rm\implies \:\boxed{\tt{ {sec}^{4}x - {tan}^{4}x = 1 + {2tan}^{2}x \: }}$

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Additional Information:-

Relationship between sides and T ratios

sin θ = Opposite Side/Hypotenuse

cos θ = Adjacent Side/Hypotenuse

tan θ = Opposite Side/Adjacent Side

sec θ = Hypotenuse/Adjacent Side

cosec θ = Hypotenuse/Opposite Side

cot θ = Adjacent Side/Opposite Side

Reciprocal Identities

cosec θ = 1/sin θ

sec θ = 1/cos θ

cot θ = 1/tan θ

sin θ = 1/cosec θ

cos θ = 1/sec θ

tan θ = 1/cot θ

Co-function Identities

sin (90°−x) = cos x

cos (90°−x) = sin x

tan (90°−x) = cot x

cot (90°−x) = tan x

sec (90°−x) = cosec x

cosec (90°−x) = sec x

Fundamental Trigonometric Identities

sin²θ + cos²θ = 1

sec²θ - tan²θ = 1

cosec²θ - cot²θ = 1