Question

plzz plzz plzz plzz plzz plzz plzz plzz plzz plzz plzz plzz plzz plzz plzz plzz plzz plzz plzz plzz plzz plzz plzz plzz plzz plzz ans very urgent my boards are from 7​

Answer 1

Step-by-step explanation:

Note: For the better understanding,I am replacing θ with A.

Now,

$Given:\frac{sinA-cosA+1}{sinA+cosA-1}$

On rationalizing we get

$=\frac{sinA-cosA+1}{sinA+cosA-1}*\frac{sinA+cosA+1}{sinA+cosA+1}$

$=\frac{sin^2A+sinAcosA+sinA-sinAcosA-cos^2A-cosA+sinA+cosA+1}{(sinA+cosA)^2-1}$

$=\frac{sin^2A+1-cos^2A+sinA+sinA}{sin^2A+cos^2A+2sinAcosA-1}$

$=\frac{sin^2A+sin^2A+2sinA}{1+2sinAcosA-1}$

$=\frac{2sin^2A+2sinA}{2sinAcosA}$

$=\frac{2sinA(1 + sinA)}{2sinAcosA}$

$=\frac{1+sinA}{cosA} * \frac{1-sinA}{1-sinA}$

$=\frac{1-sin^2A}{cosA(1-sinA)}$

$=\frac{cos^2A}{cosA(1-sinA)}$

$=\boxed{\frac{cosA}{1-sinA}}$

Hope it helps!

Answer 2

Answer:

refers to the attachment