Question

Q:-TanA=1/3, find the other trigonometrical ratios​

Answer 1

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Q:-TanA=1/3, find the other trigonometrical ratios

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Given :-

$⟹TanA= \frac{1}{3}$

∴ $⟹CotA = \frac{3}{1} = 3$

$⟹Tan A= \frac{1}{3}$

${Tan}^{2} A = \frac{1}{9}$

$⟹1 + {Tan}^{2} A = 1 + \frac{1}{9} = \frac{9 + 1}{9} = \frac{10}{9}$

∴ $⟹SecA = \sqrt{ \frac{10}{9} }$

$⟹CosA = \frac{3}{ \sqrt{10} }$

$⟹ {Cos}^{2} A= \frac{9}{10}$

$⟹1 - {Cos}^{2} A = 1 - \frac{9}{10}$

$⟹ {Sin}^{2} A = \frac{10 - 9}{10} = \frac{1}{10}$

$⟹SinA = \sqrt{ \frac{1}{10} } = \frac{1}{ \sqrt{10} }$

∴$⟹CosecA = \sqrt{10}$

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Answer 2

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Q:-TanA=1/3, find the other trigonometrical ratios

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Given :-

$=>TanA= \frac{1}{3}$

∴ $⟹CotA = \frac{3}{1} = 3$

$=>Tan A= \frac{1}{3}$

${Tan}^{2} A = \frac{1}{9}$

$=>1 + {Tan}^{2} A = 1 + \frac{1}{9} = \frac{9 + 1}{9} = \frac{10}{9}$

∴ $=>SecA = \sqrt{ \frac{10}{9} }$

$=>CosA = \frac{3}{ \sqrt{10} }$

$=> {Cos}^{2} A= \frac{9}{10}$

$=>1 - {Cos}^{2} A = 1 - \frac{9}{10}$

$=> {Sin}^{2} A = \frac{10 - 9}{10} = \frac{1}{10}$

$=>SinA = \sqrt{ \frac{1}{10} } = \frac{1}{ \sqrt{10} }$

∴$=>CosecA = \sqrt{10}$

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