Question

Solve (1-tanx)(1+sin2x)=1+tanx

Answer 1

Given :

• $(1-tanx)(1+sin2x)=1+tanx$

To find : the solution

Solution :

Now, $(1-tanx)(1+sin2x)=1+tanx$

$\Rightarrow 1+sin2x=\frac{1+tanx}{1-tanx}$

$\Rightarrow sin^{2}x+cos^{2}x+2\:sinx\:cosx=\frac{1+tanx}{1-tanx}$

[ since $sin^{2}x+cos^{2}x=1$ ]

$\Rightarrow (sinx+cosx)^{2}=\frac{1+tanx}{1-tanx}$

$\Rightarrow cos^{2}x\:(\frac{sinx}{cosx}+1)^{2}=\frac{1+tanx}{1-tanx}$

[ taking $cosx$ common ]

$\Rightarrow cos^{2}x\:(tanx+1)^{2}=\frac{1+tanx}{1-tanx}$

$\Rightarrow cos^{2}x\:(1+tanx)=\frac{1}{1-tanx}$

$\Rightarrow cos^{2}x=\frac{1}{(1-tanx)(1+tanx)}$

$\Rightarrow \frac{1}{sec^{2}x}=\frac{1}{1-tan^{2}x}$

[ since $cosx=\frac{1}{secx}$ ]

$\Rightarrow \frac{1}{1+tan^{2}x}=\frac{1}{1-tan^{2}x}$

[ since $sec^{2}x-tan^{2}x=1$ ]

$\Rightarrow \frac{1-tan^{2}x}{1+tan^{2}x}=1$

$\Rightarrow cos2x=1$

[ since $cos2x=\frac{1-tan^{2}x}{1+tan^{2}x}$ ]

$\Rightarrow cos2x=cos0$

$\Rightarrow 2x=0$

$\Rightarrow x=0$

Answer :

• Required solution is $x=0^{\circ}$