Question

if x=sin theta and y=tan theta then dy/dx

Answer 1

Answer:

Correct answer is 1/cos³t

Step-by-step explanation:

Given, x = sin t and y = tan t

On differentiating both sides w.r.t. t respectively, we get

dx/dt = cos t and dy/dt = sec² t

Therefore, dy/dt/dx/dt = sec² t/cos t = 1/cos³t

hope it will help you get

Answer 2

Question: If $x=sin\theta$ and $y=tan\theta$ then dy/dx.

Answer:

The value of $\frac{dy}{dx}$ is $sec^3\theta$.

Step-by-step explanation:

Step 1 of 2

Consider the given trigonometric functions as follows:

$x=sin\theta$ . . . . . (1)

$y=tan\theta$ . . . . . (2)

Differentiate the equation (1) with respect $\theta$.

$\frac{dx}{d\theta} =\frac{d}{d\theta} (sin\theta)$

$\frac{dx}{d\theta} =cos\theta$ (derivative of $sin\theta$ is $cos\theta$)

or,

$\frac{d\theta}{dx} =\frac{1}{cos\theta}$

Now, differentiate the equation (2) with respect $\theta$.

$\frac{dy}{d\theta} =\frac{d}{d\theta} (tan\theta)$

$\frac{dy}{d\theta} =sec^2\theta$ (derivative of $tan\theta$ is $sec^2\theta$)

Step 2 of 2

Consider the differential form as follows:

$\frac{dy}{dx}$ . . . . . (3)

Multiply and divide the expression (3) by $d\theta$.

⇒ $\frac{dy}{dx}=\frac{dy}{dx}\times \frac{d\theta}{d\theta}$

Rewrite as follows:

⇒ $\frac{dy}{dx}=\frac{d\theta}{dx}\times \frac{dy}{d\theta}$

Now, substitute the values of $\frac{d\theta}{dx}$ and $\frac{dy}{d\theta}$, we get

⇒ $\frac{dy}{dx}=\frac{1}{cos\theta}\times sec^2\theta$

⇒ $\frac{dy}{dx}=sec\theta\times sec^2\theta$ (Since $\frac{1}{cos\theta}=sec\theta$)

⇒ $\frac{dy}{dx}=sec^3\theta$

Therefore, the value of $\frac{dy}{dx}$ is $sec^3\theta$.

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