Question

(sinA-cosA+1)/(sinA+cosA-1)=2(1+cosecA)​

Answer 1

Answer:

The trigonometrical expression is sin² A + sin A - 2 cos A - 2 cos A × sin A = 0

Step-by-step explanation:

Given Trigonometrical function as :

$\frac{sin A - cos A + 1}{sin A + cos A - 1}$ = 2 (1 + cosec A)

Or, $\frac{sin A + ( 1 - cos A)}{sin A - (1 - cos A)}$ = 2 (1 + cosec A)

, Now, rationalizing

$\frac{(sin A + ( 1 - cos A)) \times (sin A + (1 - cosA))}{(sin A - (1 - cos A))\times (sin A + (1 - cos A))}$ = 2 (1 + cosec A)

Or, $\frac{(sin A + (1 - cos A))^{2}}{sin^{2} - (1-cos A)^{2}}$ = 2 ( 1 + $\dfrac{1}{\textrm sinA}$

Or, $\frac{sin^{2}A + (1 - cosA)^{2} + 2 \times sin A \times (1 - cos A)}{sin^{2}A - (1 + cos^{2}A - 2 cos A)}$ = 2 ( $\dfrac{1 + sin A}{sin A}$

Or, $\frac{sin^{2}A + 1 + cos^{2}A - 2 cos A + 2 sin A - 2 sin A cos A}{sin^{2}A - 1 - cos^{2}A +2 cos A}$ = 2 ( $\dfrac{1 + sin A}{sin A}$

Or, $\frac{sin^{2}A + 1 + cos^{2}A - 2 cos A + 2 sin A - 2 sin A cos A}{sin^{2}A - (sin^{2}A + cos^{2}A) - cos^{2}A +2 cos A}$ = 2 ( $\dfrac{1 + sin A}{sin A}$

Or, $\frac{2- 2 cos A + 2 sin A - 2 sin A cos A}{- 2cos^{2}A +2 cos A}$ =  2 ( $\dfrac{1 + sin A}{sin A}$

Or, $\frac{1-  cos A +  sin A -  sin A cos A}{- cos^{2}A + cos A}$ = 2 ( $\dfrac{1 + sin A}{sin A}$

Or, $\frac{(1-  cos A) +  sin A (1-cos A)}{cos A(1 - cos A)}$ = 2 ( $\dfrac{1 + sin A}{sin A}$

Or, $\frac{(1-  cos A) (1 + sinA)}{cos A(1 - cos A)}$ = 2 ( $\dfrac{1 + sin A}{sin A}$

Or, $\frac{(1 + sinA)}{(cos A)}$ = 2 ( $\dfrac{1 + sin A}{sin A}$

Or, sin A + sin² A = 2 cos A (1 + sin A)

Or,  sin A + sin² A = 2 cos A + 2 cos A × sin A

Or,   sin² A + sin A - 2 cos A - 2 cos A × sin A = 0

So,The trigonometrical expression is sin² A + sin A - 2 cos A - 2 cos A × sin A = 0     Answer