Question

solve this plz wrong answer will be reported​

Answer 1

Answer:

The required value is 1/6

Step-by-step explanation:

Given :

5tanθ = 4

To find :

the value of 5sinθ - 3cosθ/5sinθ + 2cosθ

Solution :

$\boxed{\sf tanA=\dfrac{sinA}{cosA}}$

  5tanθ = 4

  5(sinθ/cosθ) = 4

 5sinθ = 4cosθ

Finding the required value,

$\sf =\dfrac{5 sin \theta-3cos \theta}{5 sin \theta+2cos \theta} \\\\ \sf=\dfrac{4cos\theta-3cos \theta}{4cos \theta+2cos \theta} \\\\ \sf =\dfrac{cos \theta}{6cos \theta} \\\\ \sf = \dfrac{1}{6}$

Therefore, the value of 5sinθ - 3cosθ/5sinθ + 2cosθ is 1/6

____________________

Some Trigonometric Identities :

$\boxed{\begin{minipage}{6cm} Important Trigonometric identities :- \\ \\ \: \: 1)\:\sin^2\theta+\cos^2\theta=1 \\ \\ 2)\:\sin^2\theta= 1-\cos^2\theta \\ \\ 3)\:\cos^2\theta=1-\sin^2\theta \\ \\ 4)\:1+\cot^2\theta=\text{cosec}^2 \, \theta \\ \\5)\: \text{cosec}^2 \, \theta-\cot^2\theta =1 \\ \\ 6)\:\text{cosec}^2 \, \theta= 1+\cot^2\theta \\\ \\ 7)\:\sec^2\theta=1+\tan^2\theta \\ \\ 8)\:\sec^2\theta-\tan^2\theta=1 \\ \\ 9)\:\tan^2\theta=\sec^2\theta-1 \end{minipage}}$

Trigonometric Table :

$\Large{ \begin{tabular}{|c|c|c|c|c|c|} \cline{1-6} \theta & \sf 0^{\circ} & \sf 30^{\circ} & \sf 45^{\circ} & \sf 60^{\circ} & \sf 90^{\circ} \\ \cline{1-6} \sin & 0 & \dfrac{1}{2 } & \dfrac{1}{ \sqrt{2} } & \dfrac{ \sqrt{3}}{2} & 1 \\ \cline{1-6} \cos & 1 & \dfrac{ \sqrt{ 3 }}{2} } & \dfrac{1}{ \sqrt{2} } & \dfrac{ 1 }{ 2 } & 0 \\ \cline{1-6} \tan & 0 & \dfrac{1}{ \sqrt{3} } & 1 & \sqrt{3} & \infty \\ \cline{1-6} \cot & \infty & \sqrt{3} & 1 & \dfrac{1}{ \sqrt{3} } &0 \\ \cline{1 - 6} \sec & 1 & \dfrac{2}{ \sqrt{3}} & \sqrt{2} & 2 & \infty \\ \cline{1-6} \csc & \infty & 2 & \sqrt{2 } & \dfrac{ 2 }{ \sqrt{ 3 } } & 1 \\ \cline{1 - 6}\end{tabular}}$