Question

tan∅=4/3 then find sin∅+cos∅
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Answer 1

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tan∅=4/3 then find sin∅+cos∅

⠀⠀⠀⠀⠀⠀${ \huge{ \mathbb{ \overbrace{ \underbrace{ \purple{ANSWER}}}}}}$

$\large\underline{ \underline{ \green{ \bold{solution}}}} = >$

⠀⠀⠀⠀⠀$tan \theta = \frac{4}{3} (given) \\ \boxed{tan\theta = \frac{perpendicular}{base} }$

By Pythagoras theorem,

⠀$AC {}^{2} =AB {}^{2} +BC {}^{2} \\ AC {}^{2} = 4 {}^{2} + 3 {}^{2} \\AC {}^{2} = 16 + 9 \\ AC = \sqrt{25} \\ AC = 5$

⠀⠀⠀⠀⠀${ \boxed{sin \theta = \frac{perpendicular}{hypotenuse} }}$

⠀⠀⠀⠀⠀${ \boxed{cos \theta = \frac{base}{hypotenuse} }}$

so,

⠀⠀⠀$sin \theta + cos \theta \\ = > \frac{4}{3} + \frac{3}{5} \\ = > \frac{7}{5} \\ { \boxed{ \fbox{ \blue{sin \theta + cos \theta = \frac{7}{5} }}}}$

Answer 2

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${ \sf{Given}}{ \sf{ \implies tan \theta = \dfrac{4}{3} }} \\ { \sf{ We \: know \: that }} \\ { \sf{ tan \theta = \frac{sin \theta}{ cos \theta} }} \\ { \sf{Now}} \\ { \sf{ \implies \frac{ sin \theta}{cos \theta} = \frac{4}{3} }} \\ { \sf{ \implies 4cos \theta = 3 sin \theta }} \\ { \sf{Let \: 4 cos \theta = 3sin \theta = x}} \\ { \sf{ 3sin \theta = x}} \\ { \sf{sin \theta = \dfrac{x}{3} }} \\ { \sf{4cos \theta = x}} \\ { \sf{cos \theta = \frac{x}{4} }}$

Using sin²θ + cos²θ = 1

${ \sf{\implies { (\frac{x}{3}) }^{2} + (\frac{x}{4})^{2} }} = 1\\ { \sf{ \implies \frac{ {x}^{2} }{9} + \frac{ {x}^{2} }{16} }} = 1 \\ { \sf{ \implies \frac{16 {x}^{2} + 9 {x}^{2} }{144} }} = 1 \\ { \sf{ \implies \frac{25 {x}^{2} }{144} = 1}} \\ { \sf{ \implies { (\frac{5x}{12} )}^{2} = 1}} \\ { \sf{ \implies \frac{5x}{12} = 1}} \\ { \sf{ \implies5x = 12}} \\ { \sf{ \implies x = \frac{12}{5} }}$

Here x = 3sinθ‍ = 4cosθ

12/5 = 3sinθ

36/5 = sinθ

Assuming as Equation ( 1 )

12/5 = 4cosθ

48/5 = cosθ

Assuming as equation ( 2 )

Equation ( 1 ) + Equation ( 2 )

sinθ + cosθ = 12/5 + 48/5

sinθ + cosθ = 70/5

sinθ + cosθ = 7