Question

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

Answer 1

Answer:

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

$\huge\tt\underline\blue{Answer }$

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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

Step-by-step explanation:

Answer:

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QUESTION:−

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

\huge\tt\underline\blue{Answer }

Answer

_ _ _ _ _ _ _ _ _ _ _ _ _ _ _✍️

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Given :-

⟹TanA= \frac{1}{3}⟹TanA=

3

1

∴ ⟹CotA = \frac{3}{1} = 3⟹CotA=

1

3

=3

⟹Tan A= \frac{1}{3}⟹TanA=

3

1

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

2

A=

9

1

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

2

A=1+

9

1

=

9

9+1

=

9

10

∴ ⟹SecA = \sqrt{ \frac{10}{9} }⟹SecA=

9

10

⟹CosA = \frac{3}{ \sqrt{10} }⟹CosA=

10

3

⟹ {Cos}^{2} A= \frac{9}{10}⟹Cos

2

A=

10

9

⟹1 - {Cos}^{2} A = 1 - \frac{9}{10}⟹1−Cos

2

A=1−

10

9

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

2

A=

10

10−9

=

10

1

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

10

1

=

10

1

∴⟹CosecA = \sqrt{10}⟹CosecA=

10

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HOPE IT HELPS YOU..

_____________________

Thankyou:)