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Mnenie [13.5K]
3 years ago
10

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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