Question

$given \: sina + cosa = \dfrac{1}{5} \\ and \: 0 \leqslant a \ < \pi \\ \\ find \: tana$
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Answer 1

sin (a) + cos (a) = 1/5 → (1)

=> (sin (a) + cos (a))² = (1/5)²

=> sin² (a) + cos² (a) + 2 sin (a) cos (a) = 1/25 → (2)

=> 1 + 2 sin (a) cos (a) = 1/25

=> 2 sin (a) cos (a) = (1/25) - 1

=> 2 sin (a) cos (a) = - 24/25 → (3)

(2) - (2 · (3))

=> sin² (a) + cos² (a) + 2 sin (a) cos (a) - 2(2 sin (a) cos (a)) = 1/25 - 2 · (- 24/25)

=> sin² (a) + cos² (a) + 2 sin (a) cos (a) - 4 sin (a) cos (a) = (1/25) + (48/25)

=> sin² (a) + cos² (a) - 2 sin (a) cos (a) = 49/25

=> (sin (a) - cos (a))² = 49/25

=> sin (a) - cos (a) = ± 7/5

Let sin (a) - cos (a) = 7/5. → (4)

(1) + (4)

=> (sin (a) + cos (a)) + (sin (a) - cos (a)) = (1/5) + (7/5)

=> 2 sin (a) = 8/5

=> sin (a) = 4/5

(1) - (4)

=> (sin (a) + cos (a)) - (sin (a) - cos (a)) = (1/5) - (7/5)

=> 2 cos (a) = - 6/5

=> cos (a) = - 3/5

This implies that 'a' lies in second quadrant, since 0 ≤ a < π.

Now,

tan (a) = sin (a) / cos (a)

tan (a) = (4/5) / (- 3/5)

tan (a) = (4/5) × (- 5/3)

tan (a) = - 4/3

Let sin (a) - cos (a) = - 7/5. → (5)

(1) + (5)

=> (sin (a) + cos (a)) + (sin (a) - cos (a)) = (1/5) + (- 7/5)

=> 2 sin (a) = - 6/5

=> sin (a) = - 3/5

But since 0 ≤ a < π, 'a' lies in either first quadrant or in second quadrant, so sin (a) is non - negative in each case. Thus sin (a) can't be - 3/5.

Hence tan (a) = - 4/3.

Answer 2

Solution :-

We will be using these relations in solving :-

$\star tan \theta= \dfrac{sin\theta}{cos\theta}$

$\star cos\theta = \dfrac{1}{\sqrt{1+tan^2\theta}}$

$\star sin\theta = \dfrac{tan\theta}{\sqrt{1+tan^2\theta}}$

Now as given

$sin(a) + cos(a) = \dfrac{1}{5}$

By substituting the sin(a) and cos(a) in terms of tan(a)

$\rightarrow \dfrac{tan(a)}{\sqrt{1+tan^2(a)}} + \dfrac{1}{\sqrt{1+tan^2(a)}} = \dfrac{1}{5}$

$\rightarrow \dfrac{tan(a) + 1}{\sqrt{1+tan^2(a)}} = \dfrac{1}{5}$

Now we will take tan(a) = k

$\rightarrow \dfrac{k + 1}{\sqrt{1+k^2}} = \dfrac{1}{5}$

Now we will square both sides :-

$\rightarrow \left( \dfrac{k + 1}{\sqrt{1+k^2}} \right)^2 = \left(\dfrac{1}{5}\right)^2$

$\rightarrow \dfrac{k^2 + 2k + 1}{ 1 + k^2 } = \dfrac{1}{25}$

By cross multiplication :-

$\rightarrow 25(k^2 + 2k + 1) = 1 + k^2$

$\rightarrow 25k^2 + 50k + 25 = 1 + k^2$

$\rightarrow 24k^2 + 50k + 24 = 0$

Now solving the quadratic equation :-

$\rightarrow k = \dfrac{ -50 \pm \sqrt{(50)^2 - 4(24)(24)}}{2(24)}$

$\rightarrow k = \dfrac{ -50 \pm \sqrt{2500 - 2304}}{48}$

$\rightarrow k = \dfrac{-50 \pm \sqrt{ 196}}{48}$

$\rightarrow k = \dfrac{ -50 \pm 14}{48}$

$\rightarrow k = \dfrac{-50 + 14}{48} \: \: or \: \: \dfrac{-50 - 14}{48}$

$\rightarrow k = \dfrac{-36}{48} \: or \: \dfrac{-64}{48}$

$\rightarrow tan(a) = \dfrac{-36}{48} \: or \: \dfrac{-64}{48}$

Now as 0 < a < π

$\rightarrow tan(a) = \dfrac{-64}{48}$

$\huge{\boxed{\sf{ tan(a) = \dfrac{-4}{3}}}}$