Question

sinA-cosA+1/sinA+cosA-1=cosA/1-sinA​

Answer 1

Step-by-step explanation:

If θ be an acute angle, the values of sin θ and cos θ lies between 0 and 1 (both inclusive). The sine of the standard angles 0°, 30°, 45°, 60° ...

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Answer 2

Solution:-

L.H.S :-

$\implies$ $\sf\dfrac{SinA-CosA+1}{SinA+CosA-1}$

$\implies$ $\sf\dfrac{\dfrac{SinA-CosA+1}{CosA}}{\dfrac{SinA+CosA-1}{CosA}}$

$\implies$ $\sf\dfrac{TanA+SecA-1}{TanA-SecA+1}$

$\implies$ $\sf\dfrac{TanA+SecA-(Sec²A-Tan²A)}{TanA-SecA+1}$

$\implies$ $\sf\dfrac{TanA+SecA-(SecA+TanA)(SecA-TanA)}{TanA-SecA+1}$

$\implies$ $\sf\dfrac{TanA+SecA[1-(SecA-TanA)}{TanA-SecA+1}$

$\implies$ $\sf\dfrac{TanA+SecA(1-SecA+TanA)}{TanA-SecA+1}$

$\implies$ $\sf TanA+SecA$

$\implies$ $\sf\dfrac{SinA}{CosA}+\dfrac{1}{CosA}$

$\implies$ $\sf\dfrac{SinA+1}{CosA}$

$\implies$ $\sf\dfrac{1+SinA}{CosA}×\dfrac{1-SinA}{1-SinA}$

$\implies$ $\sf\dfrac{1-Sin²A}{CosA(1-SinA)}$

$\implies$ $\sf\dfrac{Cos²A}{CosA(1-SinA)}$

$\implies$ $\sf\dfrac{CosA}{1-SinA}$ = R.H.S