Question

Prove that:
$\frac{ \tan(theta) }{1 - cot \: theta} + \frac{cot \: theta}{1 - tan \: theta} = 1 + sec \: theta \times cosec \: theta$
20 points​

Answer 1

if u have any doubt ask to me

Answer 2

Answer:-

$\red{\bigstar}$ Value is $\large\underline{\boxed{\tt\purple{2}}}$

• Given:-

$\sf(1 + tan \theta \: + sec \theta)(1 + cot \theta \: - cosec \theta)$

★ Basic Knowledge:-

➪ $\sf\pink{tan \theta = \dfrac{sin \theta}{cos \theta}}$

➪ $\sf\pink{cot \theta = \dfrac{cos \theta}{sin \theta}}$

➪ $\sf\pink{sec \theta = \dfrac{1}{cos \theta}}$

➪ $\sf\pink{cosec \theta = \dfrac{1}{sin \theta}}$

➪ $\sf\pink{sin^2 \theta + cos^2 \theta = 1}$

• Solution:-

➪ $\sf(1 + tan \theta \: + sec \theta)(1 + cot \theta \: - cosec \theta)$

➪ $\sf \bigg(1 + \dfrac{sin \theta}{cos \theta}+ \dfrac{1}{cos \theta} \bigg) \bigg(1+ \dfrac{cos \theta}{sin \theta} - \dfrac{1}{sin \theta} \bigg)$

➪ $\sf \bigg(\dfrac{cos \theta + sin \theta + 1 }{cos \theta} \bigg) \bigg(\dfrac{sin \theta+ cos \theta - 1}{sin \theta} \bigg)$

➪ $\sf \bigg(\dfrac{(sin \theta + cos \theta - 1)(cos \theta + sin \theta + 1)}{cos \theta . sin \theta} \bigg)$

➪ $\sf \bigg(\dfrac{(sin \theta + cos \theta - 1~</p><p>➪ [tex]\sf \bigg(\dfrac{(sin \theta + cos \theta)^2 - 1^2}{sin \theta . cos \theta} \bigg)$

➪ $\sf \bigg(\dfrac{sin^2 \theta + cos^2 \theta+ 2 sin cos \theta - 1}{sin \theta . cos \theta} \bigg)$

➪ $\sf \bigg(\dfrac{1 + 2 sin cos \theta - 1}{sin \theta . cos \theta} \bigg)$

➪ $\sf \bigg(\dfrac{2 sin \theta. cos \theta}{sin \theta . cos \theta} \bigg)$

✯ $\large{\bf\green{2}}$

Therefore, the value of $\sf(1 + tan \theta \: + sec \theta)(1 + cot \theta \: - cosec \theta)$

is 2.