If sin⁡θ+cos⁡θ=15 and 0≤θ<π then tan⁡θ is

Squaring both the sides, we get1+sin⁡2θ=125⇒sin⁡2θ=−2425let t=tan⁡θ, we get 2t1+t2=−2425⇒50t+24+24t2=0⇒12t2+25t+12=0⇒(4t+3)(3t+4)=0⇒t=−4/3 (as for t=−3/4 (rejected) as if tan⁡θ=−3/4, then θ∈[π/2,π) and sin⁡θ+cos⁡θ=−1/5)

If sin⁡θ+cos⁡θ=15 and 0≤θ<π then tan⁡θ is

Squaring both the sides, we get1+sin⁡2θ=125⇒sin⁡2θ=−2425let t=tan⁡θ, we get 2t1+t2=−2425⇒50t+24+24t2=0⇒12t2+25t+12=0⇒(4t+3)(3t+4)=0⇒t=−4/3 (as for t=−3/4 (rejected) as if tan⁡θ=−3/4, then θ∈[π/2,π) and sin⁡θ+cos⁡θ=−1/5)

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If $\sin θ + \cos θ = \frac{1}{5} and 0 \leq θ < π$ then $\tan θ$ is

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$- \frac{4}{3}$

$- \frac{3}{4}$

$\frac{3}{4}$

$\frac{4}{3}$

answer is A.

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Detailed Solution

Squaring both the sides, we get
$\begin{aligned} 1 + \sin 2 θ = \frac{1}{25} \\ \Rightarrow \sin 2 θ = - \frac{24}{25} \end{aligned}$
let $t = \tan θ, we get \frac{2 t}{1 + t^{2}} = - \frac{24}{25}$
$\begin{array}{l} \Rightarrow 50 t + 24 + 24 t^{2} = 0 \\ \Rightarrow 12 t^{2} + 25 t + 12 = 0 \\ \Rightarrow (4 t + 3) (3 t + 4) = 0 \\ \Rightarrow t = - 4 / 3 (as for t = - 3 / 4 (rejected) as if \tan θ = - 3 / 4, then θ \in [π / 2, π) and \sin θ + \cos θ = - 1 / 5) \end{array}$