Question

if sintheta + cos theta = root2 the prove that tan theta + cot theta =2

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Answer 1

AnswEr :

Given that,

$\sf \: sin \: \theta + cos \: \theta = \sqrt{2}$

To Prove :

$\sf \: tan \: \theta + cot \: \theta = 2$

Consider,

$\sf \: sin \: \theta + cos \: \theta = \sqrt{2}$

Squaring on both sides of the equation,

$\longrightarrow \sf \: (sin \: \theta + cos \: \theta) = {( \sqrt{2} \: ) }^{2} \\ \\ \longrightarrow \: \sf \: {sin}^{2} \theta + cos {}^{2} \theta + 2sin \theta.cos \: \theta = 2 \\ \\ \longrightarrow \: \sf \: 2sin \: \theta.cos \: \theta = 1 \\ \\ \longrightarrow \: \boxed{\boxed{ \sf \: sin \theta.cos \theta = \dfrac{1}{2}}}$

• sin²∅ + cos²∅ = 1

Now,

$\sf \: tan \: \theta + cot \: \theta \\ \\ \dashrightarrow \: \sf \: \dfrac{sin \: \theta}{cos \: \theta} + \dfrac{cos \: \theta}{sin \: \theta} \\ \\ \dashrightarrow \: \sf \: \dfrac{ {sin}^{2} \theta + {cos}^{2} \theta }{sin \theta.cos \theta} \\ \\ \dashrightarrow \: \sf \: \dfrac{1}{( \frac{1}{2}) } \: \dashrightarrow \: 2$

Hence,ProveD

Answer 2

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

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