If 3 tan θ = 4, show that = 3.

tan θ = Let Perpendicular = 4x and Base = 3x We will find hypotenuse by using pythagoras theorem Hypotenuse2 = Perpendicular2 + Base2 Hypotenuse2 = (4x)2 + (3x)2 Hypotenuse2 = 16x2 + 9x2 Hypotenuse2 = 25x2 Hypotenuse = 5x Now sin θ = cos θ = Substituting values we get : Hence, proved that = 3.

Mathematics

If 3 tan θ = 4, show that $\Big(\dfrac{3\sin θ + 2\cos θ}{3\sin θ - 2\cos θ}\Big)$ = 3.

Trigonometrical Ratios

2 Likes

Answer

tan θ = $\dfrac{\text{perpendicular}}{\text{base}}=\dfrac{4}{3}$

Let Perpendicular = 4x and Base = 3x

We will find hypotenuse by using pythagoras theorem

Hypotenuse² = Perpendicular² + Base²

Hypotenuse² = (4x)² + (3x)²

Hypotenuse² = 16x² + 9x²

Hypotenuse² = 25x²

Hypotenuse = 5x

Now

sin θ = $\dfrac{\text{perpendicular}}{\text{hypotenuse}}= \dfrac{4x}{5x} = \dfrac{4}{5}$

cos θ = $\dfrac{\text{base}}{\text{hypotenuse}}= \dfrac{3x}{5x} = \dfrac{3}{5}$

Substituting values we get :

$\Rightarrow \dfrac{\text{3 sin θ + 2 cos θ}}{\text{3 sin θ - 2 cos θ}} \\[1em] = \dfrac{3\times\dfrac{4}{5} + 2\times\dfrac{3}{5}}{3\times\dfrac{4}{5} -2\times\dfrac{3}{5}} \\[1em] = \dfrac{\dfrac{12}{5} + \dfrac{6}{5}}{\dfrac{12}{5} -\dfrac{6}{5}}\\[1em] = \dfrac{\dfrac{12+6}{5}}{\dfrac{12-6}{5}} \\[1em] = \dfrac{\dfrac{18}{5}}{\dfrac{6}{5}} \\[1em] = \dfrac{18\times5}{5\times6}\\[1em] = \dfrac{18}{6} = 3$

Hence, proved that $\Big(\dfrac{3\sin θ + 2\cos θ}{3\sin θ - 2\cos θ}\Big)$ = 3.

Answered By

1 Like

Mathematics

If 3 tan θ = 4, show that $\Big(\dfrac{3\sin θ + 2\cos θ}{3\sin θ - 2\cos θ}\Big)$ = 3.

Trigonometrical Ratios

Answer

Related Questions

If cot θ = $\dfrac{1}{\sqrt{3}}$ , show that $\Big(\dfrac{1 - \text{cos}^2θ}{2 - \text{sin}^2θ}\Big) = \dfrac{3}{5}$ .

If sec θ = $\dfrac{13}{5}$ , show that $\dfrac{\text{2 sin θ - 3 cos θ}}{\text{4 sin θ - 9 cos θ}}$ = 3.

If cot θ = $\dfrac{q}{p}$ , show that $\Big(\dfrac{p\sin θ - q\cos θ}{p\sin θ+ q\cos θ}\Big)= \dfrac{p^2 - q^2}{p^2 + q^2}$ .

If 4 cot θ = 3, show that $\Big(\dfrac{\sin θ - \cos θ}{\sin θ+ \cos θ}\Big) = \dfrac{1}{7}$ .

Mathematics

If 3 tan θ = 4, show that (3sin⁡θ+2cos⁡θ3sin⁡θ−2cos⁡θ)\Big(\dfrac{3\sin θ + 2\cos θ}{3\sin θ - 2\cos θ}\Big)(3sinθ−2cosθ3sinθ+2cosθ​)= 3.

Trigonometrical Ratios

Answer

Related Questions

If cot θ = 13\dfrac{1}{\sqrt{3}}3​1​, show that (1−cos2θ2−sin2θ)=35\Big(\dfrac{1 - \text{cos}^2θ}{2 - \text{sin}^2θ}\Big) = \dfrac{3}{5}(2−sin2θ1−cos2θ​)=53​.

If sec θ = 135\dfrac{13}{5}513​, show that 2 sin θ - 3 cos θ4 sin θ - 9 cos θ\dfrac{\text{2 sin θ - 3 cos θ}}{\text{4 sin θ - 9 cos θ}}4 sin θ - 9 cos θ2 sin θ - 3 cos θ​ = 3.

If cot θ = qp\dfrac{q}{p}pq​, show that (psin⁡θ−qcos⁡θpsin⁡θ+qcos⁡θ)=p2−q2p2+q2\Big(\dfrac{p\sin θ - q\cos θ}{p\sin θ+ q\cos θ}\Big)= \dfrac{p^2 - q^2}{p^2 + q^2}(psinθ+qcosθpsinθ−qcosθ​)=p2+q2p2−q2​.

If 4 cot θ = 3, show that (sin⁡θ−cos⁡θsin⁡θ+cos⁡θ)=17\Big(\dfrac{\sin θ - \cos θ}{\sin θ+ \cos θ}\Big) = \dfrac{1}{7}(sinθ+cosθsinθ−cosθ​)=71​.

If 3 tan θ = 4, show that $\Big(\dfrac{3\sin θ + 2\cos θ}{3\sin θ - 2\cos θ}\Big)$ = 3.

If cot θ = $\dfrac{1}{\sqrt{3}}$ , show that $\Big(\dfrac{1 - \text{cos}^2θ}{2 - \text{sin}^2θ}\Big) = \dfrac{3}{5}$ .

If sec θ = $\dfrac{13}{5}$ , show that $\dfrac{\text{2 sin θ - 3 cos θ}}{\text{4 sin θ - 9 cos θ}}$ = 3.

If cot θ = $\dfrac{q}{p}$ , show that $\Big(\dfrac{p\sin θ - q\cos θ}{p\sin θ+ q\cos θ}\Big)= \dfrac{p^2 - q^2}{p^2 + q^2}$ .

If 4 cot θ = 3, show that $\Big(\dfrac{\sin θ - \cos θ}{\sin θ+ \cos θ}\Big) = \dfrac{1}{7}$ .