Question

Solve the question in attachment

Answer 1

Question :-

$\sf \: If \: sin\theta + cos\theta = \sqrt{2}, \: then \: prove \: that \: tan\theta + cot\theta = 2$

$\large\underline{\sf{Solution-}}$

Given that,

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

On squaring both sides, we get

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

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

can be rewritten as

$\sf \: ( 1 - {cos}^{2}\theta) + (1 - {sin}^{2}\theta ) + 2sin\theta cos\theta = 2$

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

$\sf \: - {cos}^{2}\theta - {sin}^{2}\theta + 2sin\theta cos\theta = 0$

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

can be further rewritten as

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

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

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

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

Hence,

$\bf\implies \: \: tan\theta + cot\theta = 2$

$\rule{190pt}{2pt}$

Identities uaed :-

$\sf \: {(x + y)}^{2} = {x}^{2} + {y}^{2} + 2xy$

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

$\sf \: tan\theta = \dfrac{sin\theta }{cos\theta }$

$\sf \: cot\theta = \dfrac{cos\theta }{sin\theta }$

Answer 2

Answer:

$Your \: answer \: is \: \: \: \: \: cotθ + \tanθ = 2 \: \: \: \: \: \: \: \: (proved) \\ \\ \\ \\ You \: can \: see \: in \: my \: photos \\ mark \: me \: as \: brainliest$