Question

if sintita =-5/13 and tita is in 3rd quadrant then prove that 5cot²tita+12tantita+13cosectita =0​

Answer 1

GIVEN :–

$\\ \: \: { \huge{.}} \: \: { \bold{ \sin( \theta) = - \dfrac{5}{13}}}$

$\\ \: \: { \huge{.}} \: \: { \bold{ \theta \: \: is \: \: in \: \: third \: \: quadrant .}}$

TO PROVE :–

$\\ \implies { \bold{5 \: { \cot}^{2} \theta \: + \: 12 \tan( \theta) + 13 \: cosec( \theta) = 0}}$

SOLUTION :–

• We know that –

$\\ \implies { \bold{ { \sin}^{2} ( \theta) + { \cos}^{2}( \theta) = 1 }}$

$\\ \implies { \bold{ { \cos}^{2}( \theta) = 1 - { \sin}^{2} ( \theta) }}$

$\\ \implies { \bold{ { \cos}( \theta) = \sqrt{1 - { \sin}^{2} ( \theta) }}}$

$\\ \implies { \bold{ { \cos}( \theta) = \sqrt{1 - \left({ - \dfrac{5}{13}} \right) ^{2} }}}$

$\\ \implies { \bold{ { \cos}( \theta) = \sqrt{1 - \left({ \dfrac{25}{169}} \right) }}}$

$\\ \implies { \bold{ { \cos}( \theta) = \sqrt{{ \dfrac{169 - 25}{169}} }}}$

$\\ \implies { \bold{ { \cos}( \theta) = \sqrt{{ \dfrac{144}{169}} }}}$

$\\ { \bold{ \: \: \because \: \: \theta \: \: is \: \: in \: \: third \: \: quadrant \: \: so \: \: \cos( \theta) \: \: will \: \: be \: \: negetive. }}$

$\\ \implies { \bold{ { \cos}( \theta) = \pm \: {{ \dfrac{12}{13}} }}}$

$\\ { \bold{ \: \: \therefore \: \: \cos( \theta) = - \dfrac{12}{13} }}$

• Now –

$\\ { \bold{ \: \: \longrightarrow \: \: \cot( \theta) = \dfrac{ \cos( \theta) }{ \sin( \theta) } = \dfrac{- \dfrac{12}{13} }{- \dfrac{5}{13} } = \dfrac{12}{5} }}$

$\\ { \bold{ \: \: \longrightarrow \: \: \tan( \theta) = \dfrac{ \sin( \theta) }{ \cos( \theta) } = \dfrac{- \dfrac{5}{13} }{- \dfrac{12}{13} } = \dfrac{5}{12} }}$

$\\ { \bold{ \: \: \longrightarrow \: \: cosec( \theta) = \dfrac{1}{ \sin( \theta) } = \dfrac{1}{- \dfrac{5}{13} } = - \dfrac{13}{5} }}$

• Let's take L.H.S. –

$\\ { \bold{ \: \: = \: \:5 \: { \cot}^{2} \theta \: + \: 12 \tan( \theta) + 13 \: cosec( \theta) }}$

$\\ { \bold{ \: \: = \: \:5 \: { \left( \dfrac{12}{5} \right)}^{2} + \: 12 \left( \dfrac{5}{12} \right) + 13 \: \left( - \dfrac{13}{5} \right) }}$

$\\ { \bold{ \: \: = \: \:5 \: { \left( \dfrac{144}{25} \right)} + 5 - 13 \: \left( \dfrac{13}{5} \right) }}$

$\\ { \bold{ \: \: = \: \:\: \dfrac{144}{5} + 5 - \dfrac{169}{5} }}$

$\\ { \bold{ \: \: = \: \:\: \dfrac{144 - 169}{5} + 5}}$

$\\ { \bold{ \: \: = \: \:\: \dfrac{- 25}{5} + 5}}$

$\\ { \bold{ \: \: = \: \:\: - 5+ 5}}$

$\\ { \bold{ \: \: = \:\: R.H.S. }}$

$\\ \large \: \: \: \: \: \: \: \: \: \: \: \: \: \: { \bold{ \overbrace{ \underbrace{Hence \: \: proved}}}}$

Answer 2

${\huge{\bf{\red{\underline{Solution:}}}}}$

${\bf{\blue{\underline{Given:}}}}$

$\star{\sf{ sin \theta = \frac{ - 5}{13} }}$

${\bf{\blue{\underline{To\:Prove:}}}}$

$\star{\sf{ 5 {cot}^{2} \theta + 12tan \theta + 13cosec \theta = 0 }}$

${\bf{\blue{\underline{Now:}}}}$

• Let a revolving line, starting from OX trace out an XOP = θ, where θ lies in 3rd quadrant.From P, draw PM⊥X'OX.

Now,

$: \implies{\sf{ sin \theta = \frac{ - 5}{13} }}$

$: \implies{\sf{ \frac{MP}{OP} = \frac{ - 5}{13} }}$

Let MP = -5 and OP=13

$: \implies{\sf{ {OP}^{2} = {MP}^{2} + {OM}^{2} }}$

$: \implies{\sf{ { 13}^{2} = {(-5)}^{2} + {OM}^{2} }}$

$: \implies{\sf{ 169= 25 + {OM}^{2} }}$

$: \implies{\sf{ 169-25 = {OM}^{2} }}$

$: \implies{\sf{ 144 = {OM}^{2} }}$

$: \implies{\sf{ \sqrt{144} = {OM}}}$

$: \implies{\sf{ OM = 12}}$

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Now,

• In third quadrant ,only tanθ and cotθ are positive.

$: \implies{\sf{ cos \theta = \frac{OM}{MP} = \frac{ - 12}{13} }}$

$: \implies{\sf{ tan \theta = \frac{5}{12} }}$

$: \implies{\sf{ tcot \theta = \frac{12}{5} }}$

$: \implies{\sf{ Cosec\theta = \frac{-13}{5} }}$

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$\dagger \underline{\frak{ According \: to \: the \: question : }}$

$\star{\sf{ 5 {cot}^{2} \theta + 12tan \theta + 13cosec \theta = 0 }}$

$: \implies{\sf{ 5 \times \frac{144}{25} + 12 \times \frac{5}{12} + 13 \times \frac{ - 13}{5} }}$

$: \implies{\sf{ \frac{144}{5} + 5 + 13 \times \frac{ - 13}{5} }}$

$: \implies{\sf{ \frac{144}{5} + 5 - \frac{ 169}{5} }}$

$: \implies{\sf{ \frac{144 + 25 - 169}{5} }}$

$: \implies{\sf{ \frac{-25+ 25 }{5} }}$

$: \implies{\sf{ 0 }}$

Hence proved!