Question

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

Answer 1

TO PROVE :-

• 
  `(\tan \theta + \sin \theta)/(\tan \theta - \sin \theta) = (\sec \theta + 1)/(\sec \theta - 1)`

SOLUTION :-

First of all , simplify L.H.S.

`(\tan \theta + \sin \theta)/(\tan \theta - \sin \theta)`

• Use
  `\tan \theta = (\sin \theta)/(\cos \theta)` in place of tanθ.

`=> ((\sin \theta)/(\cos \theta) + \sin \theta )/((\sin \theta)/(\cos \theta) - \sin \theta)`

• Take sinθ common from both numerator & denominator.

`=> (\sin \theta ((1)/(\cos \theta) + 1) )/(\sin \theta ((1)/(\cos \theta)- 1) )`

• Cancel the sinθ from both numerator & denominator.

`=> ((1)/(\cos \theta) +1)/((1)/(\cos \theta) -1)`

• Use
  `\sec \theta = (1)/(\cos \theta)`

`=> (\sec \theta + 1)/(\sec \theta - 1)`

∴ L.H.S. = R.H.S

Answer 2

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