Math, asked by samyuta, 20 hours ago

answer this question please​

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Answered by senboni123456
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}

Answered by sweety96088sweety
2

I hope this is helpful for u .

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