How do you verify the identify $\frac{sin θ}{csc θ} = {sin}^{2} θ$ $sintheta/csctheta=sin^2theta$ ?

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How do you verify the identify $\frac{sin θ}{csc θ} = {sin}^{2} θ$ $sintheta/csctheta=sin^2theta$ ?

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Answer 1 · 2018-03-07T13:09:08

$L H S \Rightarrow \frac{sin x}{csc x}$ $LHS=>sinx/cscx$

$\Rightarrow sin x \cdot \frac{1}{csc x}$ $=> sinx*1/cscx$

$\Rightarrow sin x \cdot sin x = {sin}^{2} x$ $=>sinx*sinx = sin^2 x$ since, $sin x = \frac{1}{csc x}$ $sinx=1/cscx$

Answer 2 · 2018-03-07T13:09:36

$please look at the following explanation.$ $"please look at the following explanation."$

Explanation:

$csc = \frac{1}{sin θ}$ $csc=1/sin theta$

$write \frac{1}{sin θ} instead csc θ in the equation \frac{sin θ}{csc θ}$ $"write "1/sin theta " instead " csc theta " in the equation " sin theta/ csc theta$

$\frac{sin θ}{\frac{1}{sin θ}} = sin θ \cdot \frac{sin θ}{1} = {sin}^{2} θ$ $sin theta/(1/sin theta)=sin theta*sin theta/1=sin^2 theta$

Answer 3 · 2018-03-07T13:14:16

See Below.

Explanation:

AllAboutCircuits

Trigonometric Ratios are of 6 kinds.

$sin x = \frac{opposite}{hypotenuse} = \frac{perpendicular}{hypotenuse}$ $sin x = "opposite"/"hypotenuse" = "perpendicular"/"hypotenuse"$

$cos x = \frac{adjacent}{hypotenuse} = \frac{base}{hypotenuse}$ $cos x = "adjacent"/"hypotenuse" = "base"/"hypotenuse"$

$tan x = \frac{opposite}{adjacent} = \frac{perpendicular}{base}$ $tan x = "opposite"/"adjacent" = "perpendicular"/"base"$

$csc x = \frac{hypotenuse}{opposite} = \frac{hypotenuse}{perpendicular}$ $csc x = "hypotenuse"/"opposite" = "hypotenuse"/"perpendicular"$

$sec x = \frac{hypotenuse}{adjacent} = \frac{hypotenuse}{base}$ $sec x = "hypotenuse"/"adjacent" = "hypotenuse"/"base"$

$cot x = \frac{adjacent}{opposite} = \frac{base}{perpendicular}$ $cot x = "adjacent"/"opposite" = "base"/"perpendicular"$

So, you can see that, $csc x$ $csc x$ is the inverse of $sin x$ $sin x$ .

So,

L.H.S = $\frac{sin x}{csc x} = sin x \cdot \frac{1}{csc x} = sin x \cdot sin x = {sin}^{2} x$ $sin x/csc x = sin x * 1/csc x = sin x * sin x = sin^2 x$ = R.H.S

Hence Proved.

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How do you verify the identify $\frac{sin θ}{csc θ} = {sin}^{2} θ$ $sintheta/csctheta=sin^2theta$ ?

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Explanation:

Explanation:

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How do you verify the identify sinθcscθ=sin2θsintheta/csctheta=sin^2theta?

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How do you verify the identify $\frac{sin θ}{csc θ} = {sin}^{2} θ$ $sintheta/csctheta=sin^2theta$ ?