Question

if tan A root 2-1 show that tan A/1+tan square​

Answer 1

Answer:

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)}

(1+tan

2

A)

tanA

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

2

2

(

2

−1)

(

2

−1)

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

after cancellation, we get

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

2(

2

)

1

/* Rationalising the denominator, we get

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

2

2

×

2

2

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

2×2

2

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

4

2

= RHS

Therefore,

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

(1+tan

2

A)

tanA

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

4

2

••••

Step-by-step explanation:

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