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

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

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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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HOPE IT HELPS YOU..

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Answer 2

Answer:

First of all remember that sin x, cos x, tan x are trigonometric “ratios”. They are the ratios of different sides of a triangle. Now let us break the ques into steps so you don’t see this as a big solution.

Consider a triangle ABC right-angled at B, and angle between AC and BC be x. To make things clear I have taken AB as altitude and BC as base such that AB/BC=tan x=Altitude/Base.(Make a triangle on paper for better understanding)

Let the common ratio between AB and BC be p. As tan x =AB/BC= 12/5 therefore AB=12p and BC=5p(when there is a ratio given you can take a common number between the two involved quantities in ratio)

Apply Pythagoras theorem : AC^2=AB^2+BC^2

AC^2=(12p)^2+(5p)^2=144p^2+25p^2=169p^2

AC=sqrt(169p^2) =13

4. Now sin x= AB/AC= 12p/13p=12/13

And cosx= BC/AC=5p/13p= 5/13

5. sin x-cos x= 12/13–5/13=7/13

The basic approach was to employ the method of finding all the other triginometric ratios when one trigonometric ratio is given to you but in this case we only needed sin x and cos x.

Hope this helps.