Question

if u really give correct answer of a question above I will mark brainliest

Answer 1

Plz mark me brainlist
sin theta-cos theta+1/sin theta+cos theta-1=1/sec theta- tan theta

by cross multiplication

sin theta- cos theta × sec theta- tan theta = 1× sin theta + cos theta

sin theta - cos theta × sec theta - tan theta=sin theta + cos theta

sec theta- tan theta =1
therefore
sin theta - cos theta/ sin theta + cos theta = 1/sec theta - tan theta

Answer 2

Here I am writing theta as A.

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

Multiply and divide by (1 - sinA)

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

$= \ \textgreater \ \frac{(sinA-cosA+1)(1 - sinA)}{(sinA+cosA-1)(1 - sinA)}$

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

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

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

$= \ \textgreater \ \frac{cosA(cosA+sinA-1)}{(sinA +cosA-1)( 1- sinA) }$

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

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

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

$= \ \textgreater \ \frac{1}{secA-tanA}$



Hope it helps!