Question

If sinx+cosx =1/5 and 0≤x<π, then tanx isplease i want it step by step

Answer 1

Answer:

$tanx \: = - \dfrac{4}{3} \: or \: - \dfrac{3}{4}$

Step-by-step explanation:

Given---->

$sinx \: + \: cosx \: = \dfrac{1}{5} \: \: \: and \: \: 0 \leqslant x \leqslant \pi$

To find ---->

$value \: of \: tanx$

Concept used --->

1)

${(a + b)}^{2} \: = {a}^{2} + {b}^{2} + 2ab$

2)

${sin}^{2}x + {cos}^{2}x = 1$

3)

$sin2x = 2 \: sinx \: cosx$

4)

$sin2x = \dfrac{2 \: tanx}{1 + {tan}^{2} x}$

Solution----> ATQ,

$sinx + cosx \: = \dfrac{1}{5}$

$squaring \: both \: sides \: we \: get$

${( \: sinx + cosx \: )}^{2} = {( \dfrac{1}{5}) }^{2}$

$= > {sin}^{2}x + {cos}^{2} x + 2 \: sinx \: cosx \: = \dfrac{1}{25}$

$= > 1 \: + sin2x \: = \dfrac{1}{25}$

$= > \: \: sin2x \: = \dfrac{1}{25} - 1$

$= > \dfrac{2tanx}{1 + {tan}^{2}x } \: = - \dfrac{24}{25}$

$= > \dfrac{tanx}{1 + {tan}^{2}x} = - \dfrac{12}{25}$

$= > 25 \: tanx \: = - 12 - 12 {tan}^{2}x$

$= > 12 {tan}^{2} x + 25tanx + 12 \: = 0$

$= > 12 {tan}^{2} x + 16 \: tanx + 9 \: tanx \: + 12 = 0$

$= > 4tanx(3tanx + 4) + 3(3tanx + 4) = 0$

$= > ( \: 3 \: tanx + 4 \: ) \: (4tanx \: + 3 \: ) \: = 0$

$if \: \: 3tanx \: + 4 \: = 0$

$tanx \: = - \dfrac{4}{3}$

$if \: \: 4tanx \: + 3 \: = 0$

$tanx \: = - \dfrac{3}{4}$