Question

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

Answer 1

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

Answer 2

Answer:

TO PROVE :-

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

SOLUTION :-

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