Math, asked by gauravshukla12nov, 7 months ago

If sin A + cos A = √2 cos A , then the value of tanA is​

Answers

Answered by Asterinn
3

Given :

 \sin(A)  +  \cos(A)  =  \sqrt{2} \cos(A)

To find :

  • tan A

Concept :

 \dfrac{\sin(A)  }{\cos(A)}  =  \tan(A)

Solution :

\implies \sin(A)  +  \cos(A)  =  \sqrt{2} \cos(A)

\implies \sin(A)    =  \sqrt{2} \cos(A) - \cos(A)

Now take out cosA common :-

\implies \sin(A)    =\cos(A) ( \sqrt{2}  - 1)

\implies \sin(A) \dfrac{1}{\cos(A)}     = ( \sqrt{2}  - 1)

\implies  \dfrac{\sin(A)}{\cos(A)}     = ( \sqrt{2}  - 1)

we know that :-

 \dfrac{\sin(A)  }{\cos(A)}  =  \tan(A)

\implies  \tan (A)     =  \sqrt{2}  - 1

Answer :

\tan (A)     =  \sqrt{2}  - 1

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\large\bf\pink{Learn\:More}

1. Cosθ = base / hypotenuse

2. cossecθ = 1/ sinθ

3. sec θ = 1/cosθ

4. Cotθ = 1/ tanθ

5. Sin²θ+ Cos²θ= 1

6. Sec²θ - tan²θ = 1

7. cosec ²θ - cot²θ = 1

8. sin(90°−θ) = cos θ

9. cos(90°−θ) = sin θ

10. tan(90°−θ) = cot θ

11. cot(90°−θ) = tan θ

12. sec(90°−θ) = cosec θ

13. cosec(90°−θ) = sec θ

14. Sin2θ = 2 sinθ cosθ

15. cos2θ = Cos²θ- Sin²θ

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\large\bf\red{Trignometric\:Table}

\begin{array}{ |c |c|c|c|c|c|} \bf\angle A &   \bf{0}^{ \circ} &  \bf{30}^{ \circ} &   \bf{45}^{ \circ}  &  \bf{60}^{ \circ} &   \bf{90}^{ \circ}  \\ \\ \rm sin A & 0 & \dfrac{1}{2}& \dfrac{1}{ \sqrt{2} } & \dfrac{ \sqrt{3} }{2} &1 \\  \\ \rm cos \: A & 1 & \dfrac{ \sqrt{3} }{2}& \dfrac{1}{ \sqrt{2} } & \dfrac{1}{2} &0 \\  \\ \rm tan A & 0 & \dfrac{1}{ \sqrt{3} }& 1 &  \sqrt{3}  & \rm Not \: De fined \\  \\ \rm cosec A &  \rm Not \: De fined & 2&  \sqrt{2}  & \dfrac{2}{ \sqrt{3} } &1 \\  \\ \rm sec A & 1 & \dfrac{2}{ \sqrt{3} }&  \sqrt{2}  & 2 & \rm Not \: De fined \\  \\ \rm cot A & \rm Not \: De fined &  \sqrt{3} & 1  &  \dfrac{1}{ \sqrt{3} } & 0 \end{array}

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