Question

If A+B = Pi radian /4 then show that​

Answer 1

See below

Explanation:

OK....this is a bit long:

(cota -1)(cotb-1) = 2 Expand L side

cota cotb - cota - cotb +1 = 2

Expand L side further and subtract 1 from both sides

cosa cosb/ sina sinb - cosa/sina - cosb/sinb = 1

Now, multiply both sides of the equation by sinasinb

cosacosb - cosasinb - cosbsina = sinasinb

Now re-arrange

-cosasinb - cosbsina = sinasinb - cosacosb <==multiply through by -1

cosasinb + cosbsina = cosacosb-sinasinb <===reduce using trig identity

sin(a+b) = cos (a+b)

given: a + b = pi/4 this becomes

sqrt(2) / 2 = sqrt (2) / 2 TRUE ! Done.

Answer 2

See below for proof.

Explanation:

Begin by using the trigonometric identity:

`\boxed{\cot(A+B) = (\cot A \cot B-1)/(\cot B + \cot A)}`

Given that A + B = π/4, then:

`\cot \left((\pi)/(4)\right) =(\cot A \cot B-1)/(\cot B + \cot A)`

Since cot(π/4) = 1, we can simplify the equation to:

`1=(\cot A \cot B-1)/(\cot B + \cot A)`

Swap sides:

`(\cot A \cot B-1)/(\cot B + \cot A)=1`

Multiply both sides by (cot B + cot A):

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

Add 2 to both sides:

`\cot A \cot B+1=\cot B + \cot A+2`

Subtract (cot B + cot A) from both sides:

`\cot A \cot B+1-\cot B - \cot A=2`

Rearrange the left side:

`\cot A \cot B - \cot A - \cot B + 1 =2`

Factor out cot A from the first two terms of the left side:

`\cot A (\cot B - 1) - \cot B + 1 =2`

Rewrite - cot B + 1 as -1(cot B - 1):

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

Factor out the common term (cot B - 1):

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

`\hrulefill`

As one calculation:

`\begin{aligned}\cot(A+B)&=(\cot A\cot B-1)/(\cot B+\cot A)\\\\\textsf{Given\;$A+B=(\pi)/(4)$}\implies \cot \left((\pi)/(4)\right)&=(\cot A\cot B-1)/(\cot B+\cot A)\\\\1&=(\cot A\cot B-1)/(\cot B+\cot A)\\\\(\cot A \cot B-1)/(\cot B+\cot A)&=1\\\\\cot A \cot B-1&=\cot B+\cot A\\\\\cot A\cot B+1&=\cot B+\cot A+2\\\\\cot A\cot B+1-\cot B-\cot A&=2\\\\\cot A \cot B-\cot A-\cot B+1&=2\\\\\cot A(\cot B-1)-\cot B+1&=2\\\\\cot A(\cot B-1)-1(\cot B-1)&=2\end{aligned}`

`(\cot A-1)(\cot B-1)&=2`