Question

sin theta +cos theta = root 3, then prove that tan theta cot theta =1

Answer 1

Step-by-step explanation:

Given Sin theta +cos theta = root 3, then prove that tan theta cot theta =1

Given sin θ + cos θ = √3

Squaring both sides we get

Sin^2 θ + cos^2 θ + 2 sin θ cos θ = 3

1 + 2 sin θ cos θ = 3

2 sin θ cos θ = 2

So sin θ cos θ = 1

Now consider  tan θ + cot θ

                  = sin θ / cos θ + cos θ / sin θ

                  = sin^2 θ + cos ^2 θ / sin θ cos θ

                   = 1 / sin θ cos θ

                    = 1/ 1

                     = 1

Answer 2

Answer:

Proved

Step-by-step explanation:

$\sf Sin \ \theta \ + \ Cos \ \theta = \sqrt{3} \\\\Square \ both \ sides, \\\\(Sin \ \theta \ + \ Cos \ \theta)^2=(\sqrt{3})^2$

Use Identity (a + b) ² =a² + b² + 2ab

$\sf Sin^2 \ \theta \ + \ Cos^2 \ \theta + 2Sin \ \theta \ Cos \ \theta=3\\\\\text{We know that} \ Sin^2 \ \theta \ + \ Cos^2 \ \theta = 1\\\\ 1 + 2Sin \ \theta \ Cos \ \theta = 3\\\\2Sin \ \theta \ Cos \ \theta = 3 - 1\\\\2Sin \ \theta \ Cos \ \theta = 2\\\\ Sin \ \theta \ Cos \ \theta = \dfrac{2}{2}\\\\Sin \ \theta \ Cos \ \theta = 1 ------------------------(i)$

$\sf LHS = tan \ \theta \ + Cot \ \theta$

      $\sf = \dfrac{Sin \ \theta}{Cos \ \theta}+ \dfrac{Cos \ \theta}{Sin \ \theta}\\\\ =\dfrac{Sin \ \theta*Sin \ \theta}{Cos \ \theta* Sin \theta} + \dfrac{Cos \ \theta*Cos \ \theta}{Sin \theta * Cos \ \theta}$

      $\sf =\dfrac{Sin^2 \ \theta}{Sin \ \theta*Cos \ \theta}+\dfrac{Cos^2 \ \theta}{Sin \ \theta*Cos \ \theta}\\\\= \dfrac{Sin^2 \ \theta + Cos^2 \ \theta}{Sin \ \theta*Cos \ \theta}\\\\= \dfrac{1}{1} \ [\bf from (i)]\\\\= 1 =RHS$

Hence proved.