If tan-1-1-2-x and tan-1-1-2x-1- then write the value of - lying in the interval 0- -2-

We have, tan α= 1/1+2^ x and tan β = 1/1+2^x+1 Now, tan(α+ β)= tan α+ tan β/1 tan α× tan β = 1/1+2^ x+1/1+2^x+1/1 1/1+2^ x×1/1+2^x+1 = 1+2^x+1+1+2^ x/(1+2^ x)( ...

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Byju's Answer

Standard XII

Mathematics

Trigonometric Ratios for Sum of Two Angles

If tanα=1/1+2...

Question

If $t a n α = \frac{1}{1 + 2^{- x}}$ and $t a n β = \frac{1}{1 + 2^{x + 1}}$ , then write the value of $α + β$ lying in the interval $(0, \frac{π}{2})$ .

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Solution

We have,
$t a n α = \frac{1}{1 + 2^{- x}}$ and $t a n β = \frac{1}{1 + 2^{x + 1}}$
Now, $t a n (α + β) = \frac{t a n α + t a n β}{1 - t a n α \times t a n β}$
$= \frac{\frac{1}{1 + 2^{- x}} + \frac{1}{1 + 2^{x + 1}}}{1 - \frac{1}{1 + 2^{- x}} \times \frac{1}{1 + 2^{x + 1}}}$
$= \frac{\frac{1 + 2^{x + 1} + 1 + 2^{- x}}{(1 + 2^{- x}) (1 + 2^{x + 1})}}{1 - \frac{1}{(1 + 2^{- x}) (1 + 2^{x + 1})}}$
$= \frac{\frac{2 + 2^{x + 1} + 1 + 2^{- x}}{(1 + 2^{x + 1} + 2^{- x} + 2)}}{1 - \frac{1}{1 + 2^{x + 1} + 2^{- x} + 2}}$
$= \frac{\frac{2 + 2^{x + 1} + 2^{- x}}{3 + 2^{x + 1} + 2^{- x}}}{1 - \frac{1}{3 + 2^{x + 1} + 2^{- x}}}$
$= \frac{\frac{2 + 2^{x + 1} + 2^{- x}}{3 + 2^{x + 1} + 2^{- x}}}{\frac{3 + 2^{x + 1} + 2^{- x} - 1}{3 + 2^{x + 1} + 2^{- x}}}$
$= \frac{2 + 2^{x + 1} + 2^{- x}}{2 + 2^{x + 1} + 2^{- x}}$
=1
$\Rightarrow t a n (α + β) = 1 = t a n (\frac{π}{4})$
$\Rightarrow t a n (α + β) = t a n (\frac{π}{4})$
$\Rightarrow α + β = \frac{π}{4}$

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If tanα=11+2−x and tanβ=11+2x+1, then write the value of α+β lying in the interval (0,π2).

If $t a n α = \frac{1}{1 + 2^{- x}}$ and $t a n β = \frac{1}{1 + 2^{x + 1}}$ , then write the value of $α + β$ lying in the interval $(0, \frac{π}{2})$ .