Question

If tan a +4=0 then 2 cot a -5cos a +sin a is equal to----.pls answer with steps

Answer 1

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$= > \tan( \theta) + 4 = 0$
$\tan( \theta) = - 4$

$so \: \: \: \: \cot( \theta) = - \frac{1}{4} \: \: \: \: - (i)$

$\tan( \theta) = \frac{ \sin( \theta) }{ \sqrt{ { \sin^{2} ( \theta) - 1 } } }$

${ \tan}^{2}( \theta) = \frac{ \sin ^{2} (\theta)}{ { \sin }^{2} (\theta) - 1 }$

${( - 4)}^{2} ( { \sin }^{2} (\theta) - 1) = { \sin }^{2}( \theta)$

$16 { \sin }^{2} ( \theta) - 16 = \sin^{2} ( \theta)$

$15 \sin^{2} (\theta) = 16$

$\sin( \theta) = \frac{4}{ \sqrt{15} } \: \: \: \: \: \: - (ii)$
so,

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

$\cos( \theta) = \sqrt{ \frac{16 - 15}{15} }$

$\cos( \theta) = \frac{ - 1}{ \sqrt{15} } \: \: \: \: - (iii)$

now,
$= > 2 \cot( \theta) - 5 \cos( \theta) + \sin( \theta)$
by eq. (i (ii & (iii

$= > 2( - \frac{1}{4} ) - 5( - \frac{1}{ \sqrt{15} } ) + \frac{4}{ \sqrt{15} }$

$= > ( - \frac{1}{2} + \frac{5}{ \sqrt{15} } + \frac{4}{ \sqrt{15} } )$

$= > (\frac{ - \sqrt{15 } + 10 + 8 }{2 \sqrt{15} } )$

$= > \frac{18 - \sqrt{15} }{2 \sqrt{15} }$

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