Question

solve please..........

Answer 1

we are given

$\frac{tanx}{secx-1}- \frac{sinx}{1+cosx}=2cot(x)$

Firstly , we will simplify left side

and make it equal to right side

Left side:

$\frac{tanx}{secx-1}- \frac{sinx}{1+cosx}$

we know that

sec(x)=1/cos(x)

$\frac{tanx}{(1/cosx)-1}- \frac{sinx}{1+cosx}$

$\frac{tanx*cosx}{1-cos(x)}- \frac{sinx}{1+cosx}$

$\frac{sin(x)}{1-cos(x)}- \frac{sinx}{1+cosx}$

now, we can take common denominator

$\frac{sin(x)(1+cosx)}{(1-cos(x))(1+cosx)}- \frac{sinx(1-cosx)}{(1+cosx)(1-cosx)}$

$\frac{sin(x)(1+cosx)}{1-cos^2(x)}- \frac{sinx(1-cosx)}{1-cos^2x}$

$\frac{sin(x)(1+cosx)}{sin^2x}- \frac{sinx(1-cosx)}{sin^2x}$

now, we can cancel sinx

$\frac{(1+cosx)}{sinx}- \frac{(1-cosx)}{sinx}$

$\frac{(1+cosx)-(1-cosx)}{sinx}$

now, we can simplify it

$\frac{1+cosx-1+cosx}{sinx}$

$\frac{2cosx}{sinx}$

so, we get

$2cot(x)$

we can see that left side is equal to right side

$\frac{tanx}{secx-1}- \frac{sinx}{1+cosx}=2cot(x)$............Answer