CBSE BOARD X, asked by Sangsktra, 1 year ago

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

Answers

Answered by ishanit
0
hey mate,
please check my answer ask if any queries
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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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Sangsktra: Can tan be negative
ishanit: tan is negative (given in question)
Sangsktra: Okay
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