Question

1. Prove that costeta/1-tanteta + sinteta/1-cotteta=sinteta+costeta​

Answer 1

Step-by-step explanation:

$\bf\underline{\underline{\red{To\:Prove:-}}}$

• $\rm \dfrac{cos \theta}{1 - tan \theta} + \dfrac{sin \theta}{1 - cot \theta} = sin \theta + cot \theta$

$\bf\underline{\underline{\green{Proof:-}}}$

$\rm LHS = \dfrac{cos \theta}{1 - tan \theta} + \dfrac{sin \theta}{1 - cot \theta}$

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

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

$\rm = \dfrac{cos \theta.cos \theta}{ cos \theta - sin \theta} + \dfrac{sin \theta.sin \theta}{ sin \theta - cos \theta}$

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

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

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

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

$\rm = \dfrac{(cos \theta - sin\theta)(cos\theta + sin\theta)}{ cos \theta - sin \theta}$

$\rm = \dfrac{ \cancel{(cos \theta - sin\theta)} \: (cos\theta + sin\theta)}{ c \cancel{cos \theta - sin \theta}}$

$\rm = sin \theta + cos \theta = RHS$

LHS = RHS

Hence Proved

$\bf\underline{\underline{\purple{Formulas\: used:-}}}$

• $\rm tan \theta = \dfrac{sin \theta}{cos \theta}$

• $\rm cot \theta = \dfrac{cos \theta}{sin \theta}$

• $\rm {a}^{2} - {b}^{2} = (a + b)(a - b)$