Question

Please prove this ASAP!!

Answer 1

Step-by-step explanation:

Hope you find this answer helpful.

Answer 2

To Prove:

$\sf \Longrightarrow sin \theta(1 + tan\theta) + cos \theta(1 + cot\theta) = sec \theta + cosec\theta$

Taking the LHS we get:

$\sf \Longrightarrow sin \theta(1 + tan\theta) + cos \theta(1 + cot\theta)$

We're asked to prove that this expression is equal to secθ + cosecθ.

We know that we can express this expression in the form of secθ and cosecθ if we express the whole expression in terms of sinθ and cosθ.

(Because 1/sinθ = cosecθ and 1/cosθ = secθ)

Using the below ratio we'll get:

⇒ tanθ = sinθ/cosθ

⇒ tanθ = cosθ/sinθ

$\sf \Longrightarrow sin \theta\Bigg[1 + \dfrac{sin \theta}{cos \theta}\Bigg] + cos \theta\Bigg[1 + \dfrac{cos \theta}{sin \theta}\Bigg]$

$\sf \Longrightarrow sin \theta\Bigg[\dfrac{cos \theta + sin \theta}{cos \theta}\Bigg] + cos \theta\Bigg[\dfrac{sin \theta + cos \theta}{sin \theta}\Bigg]$

$\sf \Longrightarrow \dfrac{sin \theta\big[cos \theta + sin \theta\big]}{cos \theta} + \dfrac{cos\theta\big[sin \theta + cos \theta\big]}{sin \theta}$

Taking LCM we get;

$\sf \Longrightarrow \dfrac{sin^2\theta\big[cos \theta + sin \theta\big] + cos^2\theta\big[sin \theta + cos \theta\big]}{sin \theta cos \theta}$

Take [cosθ + sinθ] outside the bracket since it's a common factor.

$\sf \Longrightarrow \dfrac{cos\theta + sin\theta \Big[sin^2\theta + cos^2\theta\Big]}{sin \theta cos \theta}$

We know that:

⇒ sin²θ + cos²θ = 1

$\sf \Longrightarrow \dfrac{cos\theta + sin\theta \Big[1\Big]}{sin \theta cos \theta}$

$\sf \Longrightarrow \dfrac{cos\theta + sin\theta}{sin \theta cos \theta}$

Splitting the numerator we get:

$\sf \Longrightarrow \dfrac{cos\theta}{sin \theta \ cos \theta} + \dfrac{sin\theta}{sin \theta \ cos \theta}$

$\sf \Longrightarrow \dfrac{1}{sin \theta} + \dfrac{1}{cos \theta}$

We know that:

⇒ 1/sinθ = cosecθ

⇒ 1/cosθ = secθ

Substitute these values in the previous line.

$\sf \Longrightarrow cosec \theta + sec \theta$

LHS = RHS

Hence proved.