Question

Plz ans as soon as possible

Answer 1

Step-by-step explanation:

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

• $\rm \sin \theta(1 + \tan \theta) + \cos \theta(1 + \cot \theta) = \sec \theta + \cosec \theta$

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

$\boxed{ \rm \tan \theta = \dfrac{sin \theta}{cos \theta} \: \: \: and \: \: \: cot \theta = \dfrac{cos \theta}{sin \theta} }$

$\rm = \sin \theta(1 + \tan \theta) + \cos \theta(1 + \cot \theta)$

$\rm = \sin \theta \bigg(1 + \dfrac{ \sin \theta}{cos \theta} \bigg)+ \cos \theta \bigg(1 + \dfrac{cos \theta}{sin \theta} \bigg)$

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

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

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

$\rm = \dfrac{ {sin }^{2} \theta(cos \theta + {\sin} \theta) + {cos } ^{2} \theta ( sin \theta + cos \theta) \ }{cos \theta .sin \theta}$

$\rm = \dfrac{ {sin }^{2} \theta.cos \theta + {\sin} ^{3} \theta + {cos } ^{2} \theta . sin \theta + cos ^{3} \theta) \ }{cos \theta .sin \theta}$

$\rm = \dfrac{ {sin}^{3} \theta + {cos}^{3} \theta + sin \theta \: cos \theta(sin \theta + cos \theta) }{ cos \theta.sin \theta}$

$\rm = \dfrac{ (sin \theta + cos \theta)({sin}^{2} \theta \: - sin \theta \: cos \theta+ {cos}^{2} \theta )+ sin \theta \: cos \theta(sin \theta + cos \theta) }{ cos \theta.sin \theta}$

$\rm = \dfrac{ (sin \theta + cos \theta)(1 \: - sin \theta \: cos \theta )+ sin \theta \: cos \theta(sin \theta + cos \theta) }{ cos \theta.sin \theta}$

$\rm = \dfrac{ (sin \theta + cos \theta)(1 \: - sin \theta \: cos \theta + sin \theta \: cos \theta)}{ cos \theta.sin \theta}$

$\rm = \dfrac{ (sin \theta + cos \theta)(1 )}{ cos \theta.sin \theta}$

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

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

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

$\rm = sec \theta + tan \theta =RHS$

LHS = RHS

Hence Proved