Question

pls answer this question ​

Answer 1

give me best ans and thanks plzzzzz

Answer 2

Answer:

Step-by-step explanation:

Given :

$f_n(x)=\frac{sinx}{cos3x} + \frac{sin3x}{cos3^2x} + \frac{sin3^2x}{cos3^3x} + \frac{sin3^3x}{cos3^4x} +...+\frac{sin3^{(n-1)}x}{cos3^nx}$

Then,

find :

$f_2(\frac{\pi }{4})+f_3(\frac{\pi }{4})$

Solution :

$f_2(\frac{\pi }{4}) = \frac{sin(\frac{\pi}{4})}{cos3(\frac{\pi}{4})} + \frac{sin3(\frac{\pi}{4})}{cos3^2(\frac{\pi}{4})}$

⇒ $f_2(\frac{\pi }{4}) = \frac{sin45}{cos3(45)} + \frac{sin3(45)}{cos9(45)}$

⇒ $f_2(\frac{\pi }{4}) = \frac{sin45}{cos135} + \frac{sin135}{cos405}$

⇒ $f_2(\frac{\pi }{4}) = \frac{\frac{1}{\sqrt{2} } }{-\frac{1}{\sqrt{2}}} + \frac{\frac{1}{\sqrt{2} } }{\frac{1}{\sqrt{2} } } = (-1) + 1 = 0$

$f_3(\frac{\pi }{4})$

⇒ $f_2(\frac{\pi }{4}) + \frac{sin\ 3^2(\frac{\pi }{4} )}{cos \ 3^3(\frac{\pi }{4}) }$

⇒ $0 + \frac{sin\ 9(45)}{cos \ 27(45)} = \frac{sin225}{cos1215} = \frac{-\frac{1}{\sqrt{2} } }{-\frac{1}{\sqrt{2}}} =1 = 1$

∴ $f_2(\frac{\pi }{4})+f_3(\frac{\pi }{4}) = 1$