Question

guru g's question...

plz solve this on urh note book .....

Answer 1

Here I am writing theta as A.

LHS:

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

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

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

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

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

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

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

$\frac{1+sinA}{cosA}$

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

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

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

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


RHS:

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

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

$\frac{cos^2A}{cosA - cosAsinA}$

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

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


Therefore LHS = RHS.


Hope this helps!

Answer 2

Step-by-step explanation:

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