Question

$if \: 3 \: x = sec(theta) and \: \frac{3}{x} = tan \: (theta) \: then \: 9( {x}^{2} - \frac{1}{ {x}^{2} } ) \: \:$
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Answer 1

$\green{\large\underline{\sf{Given- }}}$

$\rm :\longmapsto\:3x = sec\theta \: \: \: and \: \: \: \dfrac{3}{x} = tan\theta$

$\purple{\large\underline{\sf{To\:Find - }}}$

$\rm :\longmapsto\:9\bigg[ {x}^{2} - \dfrac{1}{ {x}^{2} } \bigg]$

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

Given that,

$\rm :\longmapsto\:3x = sec\theta$

and

$\rm :\longmapsto\:\dfrac{3}{x} = tan\theta$

We know that,

$\red{ \boxed{ \sf{ \: {sec}^{2}\theta - {tan}^{2}\theta = 1}}}$

So, on substituting the values, we get

$\rm :\longmapsto\: {(3x)}^{2} - {\bigg[\dfrac{3}{x} \bigg]}^{2} = 1$

$\rm :\longmapsto\: {9x}^{2} - \dfrac{9}{ {x}^{2} } = 1$

$\rm :\longmapsto\:9\bigg[ {x}^{2} - \dfrac{1}{ {x}^{2} } \bigg] = 1$

Hence,

$\rm :\longmapsto\:\red{ \boxed{ \sf{ \: \: \:9\bigg[ {x}^{2} - \dfrac{1}{ {x}^{2} } \bigg] \: \: = \: \: 1 \: \: }}}$

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

Answer 2

Answer:

Answer:

$The\: value \:of \:9(x^{2}-\frac{1}{x^{2}}) = 1$

Step-by-step explanation:

$\boxed{3x=sec\theta ---(1)}$

$\boxed{\frac{3}{x}=tan\theta--(2)}$

$\boxed{The \: value \: of \: 9 \: (x {}^{2} - \frac{1}{x {}^{2} } )}$

$\: \: = > \boxed{9x {}^{2} - \frac{9}{ {x}^{2} } }$

$\: \: \: = > \boxed{(3x {)}^{2} - (\frac{3}{x} ) {}^{2} }$

$= \boxed{ (secθ ) {}^{2} - ( \tanθ) {}^{2} }$

$\pink{= \boxed{\sec {}^{2} θ - \tan {}^{2} θ} = 1}$

/* By Trigonometric identity:

$\boxed {sec^{2}\theta-tan^{2}\theta=1}$

Therefore,

$The\: value \:of \:9[x^{2}-\frac{1}{x^{2}}]=1$

Hence this is Answer.