Question

If A+B=π/4,then prove that (cotA-1)(cotB-1)=2

Answer 1

your answer is in giving picture

Answer 2

If A+B=π/4 then (cotA - 1)(cotB - 1) = 2 is proved

Solution:

Given that,

$A + B = \frac{\pi}{4}$

We have to prove that:

(cotA - 1)(cotB - 1) = 2

Now let us take the given information and solve

$A+B=\frac{\pi}{4}$

Taking cot on both sides we get,

$\cot (\mathrm{A}+\mathrm{B})=\cot \left(\frac{\pi}{4}\right)$  ---- eqn 1

using the trignometric identities,

$\cot (\mathrm{A}+\mathrm{B})=\frac{\cot A \times \cot B-1}{\cot A+\cot B}$

And,

$\cot \left(\frac{\pi}{4}\right)=1$

Applying these in eqn 1, we get

$\frac{\cot A \times \cot B-1}{\cot A+\cot B}=1$

On doing cross multiplication we get,

$\cot A \cot B-1=\cot B+\cot A$

Adding 2 on both sides we get,

$\cot A \times \cot B-1+2=\cot A+\cot B+2$

$\cot A \times \cot B-\cot A-\cot B+1=2$

Now, taking common terms out we get,

$(\cot A-1)(\cot B-1)=2$

Thus proved

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