# tanA=2x(x+1)/2x+1 then find the value of sinA and cosA,cosecA

Sami Ullah
46 Points
4 years ago
$tanA=\frac{2x^2+2x}{2x+1}$
$tanA=2[\frac{x^2+x}{2x+1}]$
$tan^{2}A=4[\frac{(x^2+x)^{2}}{(2x+1)^2}]$
$tan^{2}A=4[\frac{(x^4+x^2+2x^3)}{(4x^2+4x+1)}]$
$1+tan^{2}A=1+4[\frac{(x^4+x^2+2x^3)}{(4x^2+4x+1)}]$
$sec^{2}A=\frac{4x^2+4x+1+4(x^4+x^2+2x^3)}{4x^2+4x+1}$
$sec^{2}A=\frac{4x^4+8x^3+8x^2+4x+1}{4x^2+4x+1}$
$cos^{2}A=\frac{4x^2+4x+1}{4x^4+8x^3+8x^2+4x+1}$
$cosA=\frac{2x+1}{\sqrt{4x^4+8x^3+8x^2+4x+1}}$
That is what I could do.I don’t think it can further be solved.Of it can be please notify me.Also you can now find sinA easily