Question

If cot theta+cosec theta=a,then proof that (a^2+1)cos theta=a^2-1​

Answer 1

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

Given that

$\red{\rm :\longmapsto\:cosec\theta + cot\theta = a}$

Now, Consider

$\rm :\longmapsto\:\dfrac{ {a}^{2} - 1}{ {a}^{2} + 1}$

On substituting the value of a, we get

$\rm = \: \dfrac{ {(cosec\theta + cot\theta )}^{2} - 1}{ {(cosec\theta + cot\theta )}^{2} + 1}$

We know,

$\boxed{ \tt{ \: {(x + y)}^{2} = {x}^{2} + {y}^{2} + 2xy \: }}$

So, using this identity, we get

$\rm = \: \dfrac{ {cosec}^{2}\theta + {cot}^{2}\theta + 2cosec\theta cot\theta - 1 }{{cosec}^{2}\theta + {cot}^{2}\theta + 2cosec\theta cot\theta + 1}$

can be re-arranged as

$\rm = \: \dfrac{( {cosec}^{2}\theta - 1) + {cot}^{2}\theta + 2cosec\theta cot\theta}{{cosec}^{2}\theta + ({cot}^{2}\theta + 1) + 2cosec\theta cot\theta}$

We know,

$\boxed{ \tt{ \: {cosec}^{2}\theta - {cot}^{2}\theta = 1 \: }}$

So, using this identity, we get

$\rm = \: \dfrac{{cot}^{2}\theta + {cot}^{2}\theta + 2cosec\theta cot\theta}{{cosec}^{2}\theta + {cosec}^{2}\theta + 2cosec\theta cot\theta}$

$\rm = \: \dfrac{2{cot}^{2}\theta + 2cosec\theta cot\theta}{2{cosec}^{2}\theta + 2cosec\theta cot\theta}$

$\rm = \: \dfrac{2cot\theta (cosec\theta + cot\theta )}{2cosec\theta (cosec\theta + cot\theta )}$

$\rm = \: \dfrac{cot\theta }{cosec\theta }$

$\rm = \: \dfrac{cos\theta }{sin\theta } \times sin\theta$

$\rm = \: cos\theta$

$\rm \implies\:\dfrac{ {a}^{2} - 1}{ {a}^{2} + 1} = \: cos\theta$

$\bf \implies\: {a}^{2} - 1 = ( {a}^{2} + 1)cos\theta$

Hence, Proved

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

Step-by-step explanation:

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