Question

$\rm \: \dfrac{cos \theta - sin \theta + 1}{cos \theta + sin \theta - 1} = cosec \theta + cot \theta$
solve this fast​

Answer 1

Given-

$\bf \: \dfrac{cos \theta - sin \theta + 1}{cos \theta + sin \theta - 1} = cosec \theta + cot \theta$

Concept used-

• $\bf cosec \theta = \dfrac{1}{sin \theta}$
• $\bf \: \dfrac{cos \theta}{sin \theta} = cot \theta$
• $\bf \: 1 = {cosec}^{2} \theta - {cot}^{2} \theta$

Proof-

Divide numerator and denominator of LHS by sinQ

LHS-

$\bf \cfrac{ \cfrac{cos \theta}{sin \theta} - \cfrac{sin \theta}{sin \theta} + \cfrac{1}{sin \theta} }{ \cfrac{cos \theta}{sin \theta} + \cfrac{sin \theta}{sin \theta} - \cfrac{1}{sin \theta} } \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \\ \implies \bf \cfrac{cot \theta - 1 + cosec \theta}{cot \theta + 1 - cosec \theta} \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \\ \implies \bf \cfrac{cot \theta + cosec \theta - ( {cosec}^{2} \theta - {cot}^{2} \theta) }{cot \theta - cosec \theta + 1 \theta } \\ \\ \implies \bf \cfrac{cot \theta + cosec \theta \{1 - cosec \theta + cot \theta \}}{cot \theta - cosec \theta + 1 } \\ \\ \implies \underline{ \boxed{ \bf \: LHS = cot \theta + cosec \theta}}$

RHS=cotQ+cosecQ

LHS=RHS

Hence proved