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```        more than one options may be correct
Q. for an +ve integer n, let fn(x)= {tan(x/2)}(1+secx)(1+sec2x)(1+sec4x).............(1+sec2nx). Then
(a)f2(pie/16)
(b)f3(pie/32)
(c)f4(pie/64)
(d)f5(pie/128)

```
7 years ago

147 Points
```										Dear raman
fn(x)= {tan(x/2)}(1+secx)(1+sec2x)(1+sec4x).............(1+sec2nx).
simplyfy first bracket
=tanx/2 (2cos2(x/2) /cosx))(1+sec2x)(1+sec4x).............(1+sec2nx).
=tanx (1+sec2x)(1+sec4x).............(1+sec2nx).
similery solve other bracket also
fn(x) =tan2nx
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```
7 years ago
Priyansh Goel
21 Points
```										Ans is ABCD.
```
7 years ago
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