Question

If (Sin θ + Cos θ) = √3, then find the Value of (Tan θ + Cot θ)

I need it Urgently!​

Answer 1

AnswEr :

• (Sin θ + Cos θ) = √3
• Value of (Tan θ + Cot θ)

⠀

• Let's Head to the Question Now :

$\Longrightarrow \large \sf{\sin( \theta) + \cos(\theta) = \sqrt{3} }$

⠀⠀⠀⠀⋆ Squaring Both Sides

$\Longrightarrow \large \sf{(\sin( \theta) + \cos(\theta))^{2} = (\sqrt{3})^{2} }$

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

⠀⠀⠀⠀⋆ (Sin²θ + Cos²θ) = 1

$\Longrightarrow \large \sf{1 + 2\sin( \theta)\cos(\theta) = 3}$

$\Longrightarrow \large \sf{2\sin( \theta)\cos(\theta) = 3 - 1}$

$\Longrightarrow \large \sf{ \cancel2\sin( \theta)\cos(\theta) = \cancel2}$

$\Longrightarrow \large \sf{\sin( \theta)\cos(\theta) = 1}$

⠀⠀⠀⠀⋆ Dividing Both term by Cos²θ

$\Longrightarrow \large \sf{ \dfrac{\sin( \theta) \cancel{\cos(\theta)}}{ \cancel{\cos^{2} ( \theta)} } = \dfrac{1}{ \cos^{2} ( \theta) } }$

$\Longrightarrow \large \sf{ \dfrac{ \sin( \theta) }{ \cos( \theta) } = \dfrac{1}{ \cos^{2} ( \theta) } }$

⠀⠀⠀⠀⋆ Sinθ / Cosθ = Tanθ

⠀⠀⠀⠀⋆ 1 / Cos²θ = Sec²θ

$\Longrightarrow \large \sf \tan( \theta) = \sec^{2} ( \theta)$

⠀⠀⠀⠀⋆ Sec²θ = (1 + Tan²θ)

$\Longrightarrow \large \sf \tan( \theta) = 1 + \tan^{2} ( \theta)$

$\Longrightarrow \large \sf 1 = \dfrac{1 + \tan^{2}( \theta)}{\tan( \theta)}$

$\Longrightarrow \large \sf \dfrac{1}{\tan( \theta)} + \cancel\dfrac{ \tan^{2}( \theta)}{\tan( \theta)} = 1$

$\Longrightarrow \large \sf \dfrac{1}{\tan( \theta)} +\tan( \theta)= 1$

$\Longrightarrow \large \sf \cot( \theta) +\tan( \theta)= 1$

$\Longrightarrow \boxed{\large \sf \tan( \theta) + \cot( \theta) = 1}$

⠀

჻ Hence, Value of ( Tan θ + Cot θ ) is 1.

Answer 2

Answer:

$\large\bold\red{1}$

Step-by-step explanation:

Given:

• $\sin( \theta) + \cos( \theta) = \sqrt{3}\:\:\:\:\:......(i)$

To find :

• $\tan( \theta) + \cot( \theta)$

Squaring both sides of Equation$(i),$

we get,

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

But,

we know that,

$\large\bold{{(x + y)}^{2} = {x}^{2} + 2xy + {y}^{2}}$

Therefore,

we get,

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

But,

we know the identity,

$\large\bold{{ \sin}^{2} x + { \cos }^{2} x = 1}$

Therefore,

we get,

$= > 1 + 2 \sin \theta \cos \theta = 3 \\ \\ = > 2 \sin \theta \cos \theta = 3 - 1 \\ \\ = > 2 \sin \theta \cos \theta = 2 \\ \\ = > \sin \theta \cos \theta = \frac{2}{2} \\ \\ = > \sin \theta \cos \theta = 1$

Now,

dividing both sides by $\bold{ { \cos }^{2}\theta}$,

we get,

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

But,

we know that,

$\large\bold{ \frac{ \sin(x) }{ \cos(x) } = \tan(x)} \\\\ and \\\\ \large\bold{\frac{1}{ { \cos }^{2}x } = { \sec }^{2} x}$

Therefore,

we get,

$= > \tan\theta = { \sec}^{2} \theta$

But,

we know the identity,

$\large\bold{ { \sec }^{2} x = 1 + { \tan }^{2} x}$

Therfore,

we get,

$= > \tan( \theta ) = 1 + { \tan }^{2} \theta \\ \\ = > \frac{1 + { \tan }^{2} \theta }{ \tan( \theta ) } = 1 \\ \\ = > \frac{1}{ \tan \theta} + \frac{ { \tan }^{2} \theta}{ \tan \theta } = 1$

But,

we know that,

$\large\bold{ \frac{1}{ \tan(x) } = \cot(x) }$

Therefore,

we get,

$= > \tan \theta + \cot\theta = 1$

Hence,

$\large\bold{1}$ is the required Answer.