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​If (1+tana)(1+tan 4a)=2,a ∈(0,π16)(1+\text{tan}a)(1+\text{tan}\space4a)=2, a\space∈(0,\frac{π}{16})(1+tana)(1+tan 4a)=2,a ∈(
Question

If (1+tana)(1+tan 4a)=2,a (0,π16)(1+\text{tan}a)(1+\text{tan}\space4a)=2, a\space∈(0,\frac{π}{16})​ then aa is equal to:

A.

π40\frac{π}{40}​​

B.

π60\frac{π}{60}​​

C.

π20\frac{π}{20}​​

D.

π30\frac{π}{30}​​

Correct option is C

Given:
The equation:
(1+tana)(1+tan4a)=2(1 + \tan a)(1 + \tan 4a) = 2​​
where a (0,π16)\in \left(0, \frac{\pi}{16}\right)
Concept Used:

(1 + tanA)(1 + tanB) =2

then, A + B =π4\frac{\pi}{4}​​

Solution:

(1+tana)(1+tan4a)=2(1 + \tan a)(1 + \tan 4a) = 2

aa​+ 4aa​ = π4\frac{\pi}{4}

5 aa​=π4\frac{\pi}{4}

aa​ =π20\frac{\pi}{20}(0,π16)\in \left(0, \frac{\pi}{16}\right) ​​​


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