Question

express all trigonometry ratio in terms of tan a​

Answer 1

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

We know,

$\rm :\longmapsto\: {sec}^{2}a - {tan}^{2}a = 1$

$\rm :\longmapsto\: {sec}^{2}a = {tan}^{2}a + 1$

$\bf\implies \:seca = \sqrt{1 + {tan}^{2} a}$

Now, we know

$\rm :\longmapsto\:cosa = \dfrac{1}{seca}$

$\bf\implies \:cosa = \dfrac{1}{ \sqrt{1 + {tan}^{2} a} }$

Now, We know

$\bf\implies \:cota = \dfrac{1}{tana}$

Now, We know

$\rm :\longmapsto\: {cosec}^{2}a - {cot}^{2}a = 1$

$\rm :\longmapsto\: {cosec}^{2}a = {cot}^{2}a + 1$

$\rm :\longmapsto\:coseca = \sqrt{1 + {cot}^{2} a}$

$\rm :\longmapsto\:coseca = \sqrt{1 + \dfrac{1}{ {tan}^{2} a} }$

$\rm :\longmapsto\:coseca = \sqrt{ \dfrac{ {tan}^{2}a + 1}{ {tan}^{2} a} }$

$\bf\implies \:coseca = \dfrac{ \sqrt{1 + {tan}^{2} a} }{tana}$

Now, We know that

$\rm :\longmapsto\:sina = \dfrac{1}{coseca}$

$\bf\implies \:sina = \dfrac{tana}{ \sqrt{1 + {tan}^{2} a} }$

So, we have now

$\bf\implies \:\boxed{ \tt{ \: sina = \dfrac{tana}{ \sqrt{1 + {tan}^{2} a} } \: }}$

$\bf\implies \:\boxed{ \tt{ \: cosa = \dfrac{1}{ \sqrt{1 + {tan}^{2} a} } \: }}$

$\bf\implies \:\boxed{ \tt{ \: coseca = \dfrac{ \sqrt{1 + {tan}^{2} a} }{tana} \: }}$

$\bf\implies \:\boxed{ \tt{ \: seca = \sqrt{1 + {tan}^{2} a} \: }}$

$\bf\implies \:\boxed{ \tt{ \: cota = \dfrac{1}{tana} \: }}$

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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