Question

If none of 2A, 3A and 5A are multiple of π, then Cot5ACot3A+ Cot5ACot2A- Cot3ACot2A=​

Answer 1

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

Given that

$\rm :\longmapsto\:2A, \: 3A, \: 5A \: \cancel \in \: n\pi$

Now,

$\rm :\longmapsto\:5A = 3A + 2A$

So,

$\rm :\longmapsto\:cot5A =cot( 3A + 2A)$

We know,

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

So,

$\rm :\longmapsto\:cot5A = \dfrac{cot3A \: cot2A - 1}{cot2A + cot3A}$

$\rm :\longmapsto\:cot5Acot2A + cot5Acot3A = cot3Acot2A - 1$

Hence,

$\rm :\longmapsto\:cot5Acot2A + cot5Acot3A - cot3Acot2A = - 1$

Hence, Proved

Additional Information :-

$\boxed{ \sf \: sin(x + y) = sinx \: cosy \: + \: siny \: cosx}$

$\boxed{ \sf \: sin(x - y) = sinx \: cosy \: - \: siny \: cosx}$

$\boxed{ \sf \: cos(x - y) = cosx \: cosy \: + \: siny \: sinx}$

$\boxed{ \sf \: cos(x + y) = cosx \: cosy \: - \: siny \: sinx}$

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

$\boxed{ \sf{ \: tan(x + y) = \frac{tanx + tany}{1 - tanx \: tany} }}$

$\boxed{ \sf{ \: tan(x - y) = \frac{tanx - tany}{1 + tanx \: tany} }}$

Answer 2

Answer:

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Step-by-step explanation:

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