Question

If sin theta = 4/5, find the value of 4tan theta - 5 cos theta \ sec theta + 4cot theta​

Answer 1

$\tt \purple{ \: Solution:- }$

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$\footnotesize\tt{➳ Sin \: \theta = \frac{4}{5} }$

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$\footnotesize \tt{Formula \: of \: Sin \theta = \frac{Perpendicular(P)}{Hypotenuse(H)}}$

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When we compare the given value of Sin θ with its formula we got to know that Perpendicular is 4 and Hypotenuse is 5

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$\footnotesize\tt \purple{➳ Find \: Base(BC) \: by \: using \: Pythagoras \: Theorem (PGT) }$

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$\footnotesize\tt{➳ \: AC²= AB²+ BC² }$

$\footnotesize\tt{➳ \: 5²= 4²+ BC² }$

$\footnotesize\tt{➳ \: 25= 16+ BC² }$

$\footnotesize\tt{➳ \: 25 - 16= BC² }$

$\footnotesize\tt{➳ \: 9= BC² }$

$\footnotesize\tt{➳ \: \sqrt{9} = BC }$

$\footnotesize\tt \purple{➳ \: 3 = BC }$

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$\footnotesize\tt{➳ \: Cos \: \theta = \frac{BC}{AC} }$

$\footnotesize\tt \purple{➳ \: Cos \: \theta = \frac{3}{5} }$

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$\footnotesize\tt{➳ \: tan \: \theta = \frac{AB}{BC} }$

$\footnotesize\tt \purple{➳ \: tan \: \theta = \frac{4}{3} }$

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$\footnotesize\tt{➳ \: cot \: \theta = \frac{BC}{AB}}$

$\footnotesize\tt \purple{➳ \: cot \: \theta = \frac{3}{4} }$

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$\footnotesize\tt{➳ \: Sec \: \theta = \frac{AC}{BC}}$

$\footnotesize\tt \purple{➳ \: Sec \: \theta = \frac{5}{3}}$

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$\footnotesize\tt{➳ \: Now, \frac{4tan \: \theta \: - \: 5 cos \: \theta}{sec \: \theta \: + \: 4cot \: \theta}}$

$\footnotesize➳ \: \tt{ \frac{4 \times \frac{4}{3} - \: 5 \times \frac{3}{5} }{ \frac{5}{3} \: + \: 4 \times \frac{3}{4} }}$

$\footnotesize➳ \: \tt{ \frac{4 \times \frac{4}{3} - \: \cancel 5 \times \frac{3}{ \cancel5} }{ \frac{5}{3} \: + \: \cancel4 \times \frac{3}{ \cancel4} }}$

$\footnotesize➳ \: \tt{ \frac{ \frac{16}{3} - 3}{ \frac{5}{3} + 3 }}$

$\footnotesize➳ \: \tt{ \frac{ \frac{16 - 9}{3} }{ \frac{5 + 9}{3} }}$

$\footnotesize➳ \: \tt{ \frac{ \frac{7}{3} }{ \frac{14}{3} }}$

$\footnotesize➳ \: \tt{ \frac{ \frac{7}{ \cancel3} }{ \frac{14}{ \cancel3} }}$

$\footnotesize➳ \: \tt{ \frac{7}{ 14} }$

$\footnotesize➳ \: \tt{ \cancel\frac{7}{ 14} }$

$\footnotesize➳ \: \tt \purple{ \frac{1}{2} }$

Answer 2

Answer:

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