Question

$\huge\underline\mathfrak\blue{Question\:-}$


If sinØ + cosØ = √3, prove that tanØ + cotØ = 1


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Answer 1

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Answer 2

To Prove:

tanØ + cotØ = 1 , if sinØ + cosØ = √3

Proof:

Given,

sinØ + cosØ = √3

Squaring on both sides,

we get

${(sin (\phi) + cos (\phi) )}^{2}= {(\sqrt{3})}^{2}$

But,

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

Therefore,

$=>{sin}^{2}(\phi)+{cos}^{2}(\phi) +2sin(\phi)cos(\phi)=3$

But,

${sin}^{2} (\phi) + {cos}^{2} (\phi) =1$

Therefore,

$=> 1 +2sin (\phi) cos (\phi) =3$

$=>2sin (\phi) cos (\phi)=3-1=2$

$=>sin(\phi)cos(\phi)=\frac{2}{2}=1$

On Cross multiplication,

we get,

$=>\frac{1}{sin(\phi)cos(\phi)}=1$

But, we can replace

${sin}^{2}(\phi)+{cos}^{2}(\phi)=1$

Therefore,

We get,

$=>\frac{{sin}^{2}(\phi)+{cos}^{2}(\phi)}{sin(\phi)cos(\phi)}=1$

$=>\frac{{sin}^{2}(\phi)}{sin(\phi)cos(\phi)}+\frac{{cos}^{2}(\phi)}{sin(\phi)cos(\phi)}=1$

$=>tan(\phi)+ cot(\phi) = 1$

Hence, proved