Question

Prove that
1+cot theta/sin theta + 1+tan theta/cos theta = cosec theta sec² theta+sec theta cosec² theta

(using LHS/RHS method)



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Answer 1

$\large\underline{\sf{To\:prove - }}$

$\sf \: \dfrac{1 + cot\theta}{sin\theta} + \dfrac{1 + tan\theta}{cos\theta} = cosec\theta \: {sec}^{2}\theta + sec\theta {cosec}^{2}\theta$

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

Consider, LHS

$\rm :\longmapsto\: \dfrac{1 + cot\theta}{sin\theta} + \dfrac{1 + tan\theta}{cos\theta}$

We know,

$\boxed{ \tt{ \: cotx = \frac{cosx}{sinx} \: \: and \: \: tanx = \frac{sinx}{cosx} \: }}$

So, on substituting these Identities, we get

$\rm \: = \:\dfrac{1 + \dfrac{cos\theta}{sin\theta} }{sin\theta} + \dfrac{1 + \dfrac{sin\theta}{cos\theta} }{cos\theta}$

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

$\rm \: = \:\dfrac{sin\theta + cos\theta}{ {sin}^{2}\theta } + \dfrac{sin\theta + cos\theta}{ {cos}^{2} \theta}$

$\rm \: = \:(sin\theta + cos\theta)\bigg[\dfrac{1}{ {sin}^{2} \theta} + \dfrac{1}{ {cos}^{2} \theta} \bigg]$

$\rm \: = \:(sin\theta + cos\theta)\bigg[\dfrac{ {cos}^{2} \theta + {sin}^{2} \theta}{ {sin}^{2} \theta {cos}^{2}\theta } \bigg]$

We know,

$\rm :\longmapsto\:\boxed{ \tt{ \: {sin}^{2}x + {cos}^{2}x = 1 \: }}$

So, using this identity, we get

$\rm \: = \:(sin\theta + cos\theta)\bigg[\dfrac{1}{ {sin}^{2} \theta \: {cos}^{2}\theta} \bigg]$

$\rm \: = \:\dfrac{sin\theta + cos\theta}{ {sin}^{2} \theta {cos}^{2} \theta}$

$\rm \: = \:\dfrac{sin\theta }{ {sin}^{2} \theta {cos}^{2} \theta} + \dfrac{cos\theta }{ {sin}^{2} \theta {cos}^{2} \theta}$

$\rm \: = \:\dfrac{1}{sin\theta {cos}^{2} \theta} + \dfrac{1}{cos\theta {sin}^{2} \theta}$

We know,

$\boxed{ \tt{ \: \frac{1}{sinx} = cosecx \: \: \: and \: \: \: \frac{1}{cosx} = secx \: }}$

So, using this, we get

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

Hence,

$\red{ \boxed{\sf \: \dfrac{1 + cot\theta}{sin\theta} + \dfrac{1 + tan\theta}{cos\theta} = cosec\theta \: {sec}^{2}\theta + sec\theta {cosec}^{2}\theta}}$

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