Question

Prove that
tan theta/1+cot theta+cot theta/1+tan theta=cosec theta×sec theta-1

Answer 1

Prove that:

$\Longrightarrow \sf \dfrac{tan\theta}{1 + cot \theta} + \dfrac{cot \theta}{1 + tan \theta} = cosec \theta \times sec \theta - 1$

Solution:

The RHS is in terms of cosecθ and secθ, so let's try to express the LHS in terms of cosecθ and secθ as well.

$\Longrightarrow \sf \dfrac{tan\theta}{1 + cot \theta} + \dfrac{cot \theta}{1 + tan \theta}$

Lets use;

➝ tanθ = sinθ/cosθ

➝ cotθ = cosθ/sinθ

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

Take LCM;

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

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

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

$\Longrightarrow \sf \dfrac{1}{sin\theta + cos\theta} \Bigg[\dfrac{sin^{2} \theta}{cos\theta} \ + \ \dfrac{cos^{2} \theta}{sin\theta}\Bigg]$

$\Longrightarrow \sf \dfrac{1}{sin\theta + cos\theta} \Bigg[\dfrac{sin^{3}\theta + cos^{3}\theta}{cos\theta \ sin\theta}\Bigg]$

Use a³ + b³ = (a + b)(a² - ab + b²) where; a ➝ sinθ & b ➝ cosθ

$\Longrightarrow \sf \dfrac{1}{sin\theta + cos\theta} \Bigg[\dfrac{\big(sin\theta + cos\theta\big) \big(sin^{2}\theta - sin\theta \ cos\theta + cos^{2}\theta\big)}{cos\theta \ sin\theta}\Bigg]$

Use sin²θ + cos²θ = 1.

$\Longrightarrow \sf \dfrac{1}{sin\theta + cos\theta} \Bigg[\dfrac{\big(sin\theta + cos\theta\big) \big(1 - sin\theta \ cos\theta\big)}{cos\theta \ sin\theta}\Bigg]$

Cancel (sinθ + cosθ).

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

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

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

$\Longrightarrow \sf \dfrac{1}{cos\theta} \times \dfrac{1}{sin\theta} - 1$

Using the below values we get:

➝ 1/cosθ = secθ

➝ 1/sinθ = cosecθ

$\Longrightarrow \sf sec\theta \times cosec\theta - 1$

$\Longrightarrow \sf cosec\theta \times sec\theta - 1$

LHS = RHS

Hence proved.