Question

if a+b =pie/4,then prove that 1) (1+tana).(1+tanb)=2 2) ( cota-1).(cotb-1)​

Answer 1

Appropriate Question :-

If

$\rm :\longmapsto\:a + b = \dfrac{\pi}{4}$

Prove that,

$\rm :\longmapsto\:(i) \: \: (1 + tana)(1 + tanb) = 2$

$\rm :\longmapsto\:(ii) \: \: (cota - 1)(cotb - 1) = 2$

$\red{\large\underline{\sf{Solution-i}}}$

Given that,

$\rm :\longmapsto\:a + b = \dfrac{\pi}{4}$

So,

$\rm :\longmapsto\:tan(a + b) = tan\bigg[\dfrac{\pi}{4} \bigg]$

$\rm :\longmapsto\:\dfrac{tana + tanb}{1 - tana \: tanb} = 1$

$\rm :\longmapsto\:tana + tanb = 1 - tana \: tanb$

$\rm :\longmapsto\:tana + tanb + tana \: tanb = 1$

Adding 1 on both sides, we get

$\rm :\longmapsto\:1 + tana + tanb + tana \: tanb = 1 + 1$

$\rm :\longmapsto\:(1 + tana) + tanb(1 + tana) = 2$

$\rm :\longmapsto\:(1 + tana)(1 + tanb) = 2$

Hence, Proved

$\red{\large\underline{\sf{Solution-ii}}}$

Given that,

$\rm :\longmapsto\:a + b = \dfrac{\pi}{4}$

So,

$\rm :\longmapsto\:cot(a + b) = cot\bigg[\dfrac{\pi}{4} \bigg]$

$\rm :\longmapsto\:\dfrac{cotacotb - 1}{cotb + cota} = 1$

$\rm :\longmapsto\:cotacotb - 1 = cotb + cota$

$\rm :\longmapsto\:cotacotb - cotb - cota = 1$

Adding 1 on both sides, we get

$\rm :\longmapsto\:cotacotb - cotb - cota + 1 = 1 + 1$

$\rm :\longmapsto\:cotb(cota - 1) - 1(cota - 1) = 2$

$\rm :\longmapsto\:(cota - 1)(cotb - 1) = 2$

Hence, Proved

Additional Information :-

$\boxed{ \tt{ \: sin(x + y) = sinxcosy + sinycosx}}$

$\boxed{ \tt{ \: sin(x - y) = sinxcosy - sinycosx}}$

$\boxed{ \tt{ \: cos(x + y) = cosxcosy - sinxsiny}}$

$\boxed{ \tt{ \: cos(x - y) = cosxcosy + sinxsiny}}$

$\boxed{ \tt{ \:tan(x + y) = \dfrac{tanx + tany}{1 - tanxtany}}}$

$\boxed{ \tt{ \:tan(x - y) = \dfrac{tanx - tany}{1 + tanxtany}}}$

$\boxed{ \tt{ \: cot(x + y) = \frac{cotxcoty - 1}{coty + cotx}}}$

$\boxed{ \tt{ \: cot(x - y) = \frac{cotxcoty + 1}{coty - cotx}}}$

Answer 2

Answer:

$Given that, \\ </p><p></p><p>\rm :\longmapsto\:a + b = \dfrac{\pi}{4} \\ </p><p></p><p>So,</p><p></p><p>\rm :\longmapsto\:tan(a + b) = tan\bigg[\dfrac{\pi}{4} \bigg] \\ </p><p></p><p>\rm :\longmapsto\:\dfrac{tana + tanb}{1 - tana \: tanb} = 1 \\ </p><p></p><p>\rm :\longmapsto\:tana + tanb = 1 - tana \: tanb</p><p> \\ </p><p>\rm :\longmapsto\:tana + tanb + tana \: tanb = 1 \\ </p><p></p><p>Adding 1 on both sides, we get \\ </p><p></p><p>\rm :\longmapsto\:1 + tana + tanb + tana \: tanb = 1 +1 \\ </p><p></p><p>\rm :\longmapsto\:(1 + tana) + tanb(1 + tana) = 2 \\ </p><p></p><p>\rm :\longmapsto\:(1 + tana)(1 + tanb) = 2 \\ </p><p></p><p>Hence, Proved</p><p></p><p>$