Question

give step by step answer for the attachment above​

Answer 1

QUESTION:

PROVE:

$\displaystyle \frac{ \tan \theta}{ 1 - \cot \theta} + \frac{ \cot\theta}{ 1 - \tan \theta} = \sec \theta \csc \theta + 1$

GIVEN:

$\displaystyle \frac{ \tan \theta}{ 1 - \cot \theta} + \frac{ \cot\theta}{ 1 - \tan \theta} = \sec \theta \csc \theta + 1$

TO PROVE:

$\displaystyle \frac{ \tan \theta}{ 1 - \cot \theta} + \frac{ \cot\theta}{ 1 - \tan \theta} = \sec \theta \csc \theta + 1$

FORMUALE:

★ tan ∅ = sin ∅ / cos ∅

★ cot ∅ = cos ∅ / sin ∅

★ sin² ∅ + cos² ∅ = 1

★ X³ - Y³ = ( X - Y)(X² + XY + Y²)

★ 1 / cos ∅ = sec ∅

★ 1 / sin ∅ = cosec ∅

PROOF:

$\displaystyle \frac{ \tan \theta}{ 1 - \cot \theta} + \frac{ \cot\theta}{ 1 - \tan \theta} = \sec \theta \csc \theta + 1 \\ \\ \\ \frac{\frac{ \sin \theta}{ \cos \theta} }{1 - \frac{ \cos \theta}{ \sin\theta} } + \frac{\frac{ \cos \theta}{ \sin \theta} }{1 - \frac{ \sin \theta}{ \cos \theta} } \\ \\ \\ \frac{\frac{ \sin \theta}{ \cos \theta} }{\frac{ \sin \theta - \cos \theta}{ \sin\theta} }+ \frac{\frac{ \cos \theta}{ \sin \theta} }{ \frac{ \cos \theta - \sin \theta}{ \cos \theta} } \\ \\ \\ \frac{ { \sin}^{2} \theta}{ \cos \theta( \sin \theta - \cos \theta)} + \frac{ { \cos}^{2} \theta}{ \sin \theta( \cos\theta - \sin\theta)} \\ \\ \\ \frac{ { \sin}^{2} \theta}{ \cos \theta( \sin \theta - \cos \theta)} - \frac{ { \cos}^{2} \theta}{ \sin \theta( \sin\theta - \cos\theta)} \\ \\ \\ \frac{ { \sin}^{3} \theta - { \cos}^{3} \theta}{ \sin \theta \cos \theta( \sin \theta - \cos \theta)} \\ \\ \\ \frac{ (\sin \theta - \cos \theta)( \sin^2 \theta + \sin \theta \cos \theta + \cos^2 \theta)}{\sin \theta \cos \theta( \sin \theta - \cos \theta)} \\ \\ \\ \frac{( \sin^2\theta +\cos^2 \theta + \sin \theta \cos \theta )}{\sin \theta \cos \theta} \\ \\ \\ \frac{1 + \sin \theta \cos \theta}{\sin \theta \cos \theta} \\ \\ \\ \frac{1}{\sin \theta \cos \theta} + \frac{\sin \theta \cos \theta}{\sin \theta \cos \theta} \\ \\ \sec \theta \csc \theta + 1 \\ \\ \\ \fbox{ \bf{HENCE \: PROVED}}$

Answer 2

Step-by-step explanation:

⭐AnswEr:⭐

consider LHS:

$\frac{ \tan(θ) }{1 - \cot(θ) } + \frac{ \cot(θ) }{1 - tanθ}$

$= \frac{tanθ}{1 - \frac{1}{tanθ} } + \frac{ \frac{1}{tanθ} }{1 - tanθ}$

$= \frac{ \frac{tanθ}{tanθ - 1} }{tanθ} + \frac{1}{tanθ(1 - tanθ)}$

$= \frac{ {tan}^{2}θ }{tanθ - 1} + \frac{1}{tanθ(1 - tanθ)}$

$= \frac{ - tan ^{2} θ}{1 - tanθ} + \frac{1}{tanθ(1 - tanθ)}$

$= \frac{ { - tan}^{3} θ + 1}{tanθ(1 - tanθ)}$

$= \frac{1 - tan ^{3}θ }{tanθ(1 - tanθ)}$

$= \frac{(1 - tanθ)(1 + tan ^{2}θ + tanθ) }{tanθ(1 - tanθ)}$

$= \frac{ { \sec }^{2}θ }{ \tanθ } + \frac{tanθ}{tanθ}$

$= {sec}^{2} θcotθ = 1$

$= \frac{1}{ {cos}^{2} θ} \times \frac{cosθ}{sinθ} + 1$

$= \frac{1}{cososinθ} + 1$

$= 1 + secθcosecθ$

HENCE PROVED