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3 years ago

Tantheta+sintheta/tantheta-sintheta=sectheta+1/sectheta-1

Mathematics

2 answers:

d1i1m1o1n [39]3 years ago

7 0

Step-by-step explanation:

sin(theta)÷costheta+sintheta/sintheta÷COStheta- sintheta

sintheta+sintheta×costheta/sintheta-sintheta-costheta

sintheta(1+Costheta)/sintheta(1-costheta)

1+1÷Sectheta/1-1÷Sectheta

sectheta+1÷sectheta/sectheta-1÷sectheta

sectheta+1/sectheta-1

erastovalidia [21]3 years ago

3 0

Answer:

<u>TO PROVE :-</u>

$\frac{\tan \theta + \sin \theta}{\tan \theta - \sin \theta} = \frac{\sec \theta + 1}{\sec \theta - 1}$

<u>SOLUTION :-</u>

First of all , simplify L.H.S.

$\frac{\tan \theta + \sin \theta}{\tan \theta - \sin \theta}$

Use $\tan \theta = \frac{\sin \theta}{\cos \theta}$ in place of tanθ.

$=> \frac{\frac{\sin \theta}{\cos \theta} + \sin \theta }{\frac{\sin \theta}{\cos \theta} - \sin \theta}$

Take sinθ common from both numerator & denominator.

$=> \frac{\sin \theta (\frac{1}{\cos \theta} + 1) }{\sin \theta (\frac{1}{\cos \theta}- 1) }$

Cancel the sinθ from both numerator & denominator.

$=> \frac{\frac{1}{\cos \theta} +1}{\frac{1}{\cos \theta} -1}$

Use $\sec \theta = \frac{1}{\cos \theta}$

$=> \frac{\sec \theta + 1}{\sec \theta - 1}$

∴ L.H.S. = R.H.S

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Rewrite the following integral in spherical coordinates.

lora16 [44]

In cylindrical coordinates, we have $r^2=x^2+y^2$ , so that

$z = \pm \sqrt{2-r^2} = \pm \sqrt{2-x^2-y^2}$

correspond to the upper and lower halves of a sphere with radius $\sqrt2$ . In spherical coordinates, this sphere is $\rho=\sqrt2$ .

$1 \le r \le \sqrt2$ means our region is between two cylinders with radius 1 and $\sqrt2$ . In spherical coordinates, the inner cylinder has equation

$x^2+y^2 = 1 \implies \rho^2\cos^2(\theta) \sin^2(\phi) + \rho^2\sin^2(\theta) \sin^2(\phi) = \rho^2 \sin^2(\phi) = 1 \\\\ \implies \rho^2 = \csc^2(\phi) \\\\ \implies \rho = \csc(\phi)$

This cylinder meets the sphere when

$x^2 + y^2 + z^2 = 1 + z^2 = 2 \implies z^2 = 1 \\\\ \implies \rho^2 \cos^2(\phi) = 1 \\\\ \implies \rho^2 = \sec^2(\phi) \\\\ \implies \rho = \sec(\phi)$

which occurs at

$\csc(\phi) = \sec(\phi) \implies \tan(\phi) = 1 \implies \phi = \dfrac\pi4+n\pi$

where $n\in\Bbb Z$ . Then $\frac\pi4\le\phi\le\frac{3\pi}4$ .

The volume element transforms to

$dx\,dy\,dz = r\,dr\,d\theta\,dz = \rho^2 \sin(\phi) \, d\rho \, d\theta \, d\phi$

Putting everything together, we have

$\displaystyle \int_0^{2\pi} \int_1^{\sqrt2} \int_{-\sqrt{2-r^2}}^{\sqrt{2-r^2}} r \, dz \, dr \, d\theta = \boxed{\int_0^{2\pi} \int_{\pi/4}^{3\pi/4} \int_{\csc(\phi)}^{\sqrt2} \rho^2 \sin(\phi) \, d\rho \, d\phi \, d\theta} = \frac{4\pi}3$