Question

answer this question please​

Answer 1

Answer:

Step-by-step explanation:

We have,

$\tt{f\left(\dfrac{2\,tan(x)}{1+tan^2(x)}\right)=\dfrac{\left(cos(2x)+1\right)\left(sec^2(x)+2\,tan(x)\right)}{2}}$

$\tt{\implies\,f\left(sin(2x)\right)=\dfrac{\left(cos(2x)+1\right)\left(\dfrac{1}{cos^2(x)}+2\,\dfrac{sin(x)}{cos(x)}\right)}{2}}$

$\tt{\implies\,f\left(sin(2x)\right)=\dfrac{\left(cos(2x)+1\right)\left(\dfrac{1+2sin(x)\,cos(x)}{cos^2(x)}\right)}{2}}$

$\tt{\implies\,f\left(sin(2x)\right)=\dfrac{\left(cos(2x)+1\right)\left(1+2sin(x)\,cos(x)\right)}{2\,cos^2(x)}}$

$\tt{\implies\,f\left(sin(2x)\right)=\dfrac{\left(cos(2x)+1\right)\left(1+2sin(x)\,cos(x)\right)}{1+cos(2x)}}$

$\tt{\implies\,f\left(sin(2x)\right)=1+sin(2x)}$

So,

$\tt{\implies\,f(k)=1+k}$

$\tt{\implies\,f(4)=1+4=5}$

Answer 2

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