Question

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Answer 1

$\star\:\:\:\bf\large\underline\blue{Question:-}$

• If $\sin \theta + \cos \theta = \sqrt{3}$, prove that $\tan \theta + \cot \theta = 1$

$\star\:\:\:\bf\large\underline\blue{Proof:-}$

Given that,

$\sin \theta + \cos \theta = \sqrt{3}$

Squaring both sides, we get,

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

$\implies { \sin }^{2} \theta + \cos^{2} \theta + 2 \sin \theta \cos \theta = 3$

Now, we know that,

${ \sin}^{2} \theta + { \cos}^{2} \theta = 1$

Therefore,

${ \sin }^{2} \theta + \cos^{2} \theta + 2 \sin \theta \cos \theta = 3$

$\implies1 + 2 \sin \theta \cos \theta = 3$

$\implies2 \sin \theta \cos \theta = 2$

$\implies \sin \theta \cos \theta = 1$

$\bf\underline\blue{LHS}$

$\tan \theta + \cot \theta$

$= \frac{ \sin \theta }{ \cos \theta} + \frac{ \cos \theta }{ \sin \theta}$

$= \frac{( \sin \theta)^{2} + { (\cos \theta)}^{2} }{ \sin \theta \cos \theta }$

Now, putting the values, we get,

$\frac{( \sin \theta)^{2} + { (\cos \theta)}^{2} }{ \sin \theta \cos \theta }$

$= \frac{1}{1}$

$= 1$

$\bf\underline\blue{RHS}$

$= 1$

Therefore,

$\bf\large\underline\blue{LHS=RHS}$

$\star\:\:\:\bf\large\underline\blue{Hence \:Proved. }$