Question

If tan A = √2 – 1, show that tan A/1+tan^2 A = root2/4

Answer 1

Heya User,

Hope you find your solution

Answer 2

Solution:

Given tanA= √2-1-----(1)

i) 1+ tan²A

= 1+(√2-1)²

= 1+(√2)²+1²-2*√2*1

/* (a-b)² = a²-2ab+b² * /

= = 1+2+1-2√2

= 4-2√2

= 2*2-2√2

= 2*√2*√2-2√2

Take 2√2 common, we get

= 2√2(√2-1) ----(2)

Now ,

LHS=

$\frac{tanA}{(1+tan^{2}A)}$

= $\frac{(\sqrt{2}-1)}{2\sqrt{2}(\sqrt{2}-1)}$

/* from (1) and (2) */

after cancellation, we get

= $\frac{1}{2(\sqrt{2})}$

/* Rationalising the denominator, we get

= $\frac{\sqrt{2}}{2\sqrt{2}\times\sqrt{2}}$

= $\frac{\sqrt{2}}{2\times2}$

= $\frac{\sqrt{2}}{4}$

= RHS

Therefore,

$\frac{tanA}{(1+tan^{2}A)}$ = $\frac{\sqrt{2}}{4}$

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