1.

If `sin x + cos x = 1/5`, then tan 2x is

Answer» `sinx+cosx=1/5`
we can replace `cosx=sqrt(1-sin^2(x))`
=>`sqrt(1-sin^2(x))=1/5-sinx`
squaring on both sides
`1-sin^(2)x=1/25+sin^(2)x-2/5sinx`
on solving we get
=>`50sin^(2)x-10sinx-24=0`
=>`50sin^(2)x-40sinx+30sinx-24=0`
=>`(5sinx-4)(10sinx+6)=0`
=> thus, `sinx=4/5 or -3/5` thus `cosx=-3/5 or 4/5`
thus `tanx=-4/3 or -3/4`
`tan2x=(2tanx)/(1-tan^(2)x)`
substituting values of `tanx` we get `tan2x=24/7 or -24/7`


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