Question

If tan a = k tanp, prove that (k+ 1) sin (a - b) = (k - 1) sin(a+b).​

Answer 1

Appropriate Question :-

If tana = k tanb, prove that (k + 1) sin(a - b) = (k - 1) sin(a + b).

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

Given that,

$\rm :\longmapsto\:tana = k \: tanb$

$\rm :\longmapsto\:\dfrac{tana}{tanb} = k$

$\rm :\longmapsto\:\dfrac{sina \: cosb}{cosa \: sinb} = \dfrac{k}{1}$

$\: \: \: \: \red{\bigg \{ \because \: tanx = \dfrac{sinx}{cosx} \bigg \}}$

On applying Componendo and Dividendo, we get

$\rm :\longmapsto\:\dfrac{sina \: cosb \: + \: cosa \: sinb}{sina \: cosb \: - \: cosa \: sinb} = \dfrac{k + 1}{k - 1}$

$\: \: \: \: \: \: \red{\bigg \{ \because \:\dfrac{a}{b} = \dfrac{c}{d} \: then \: \dfrac{a + b}{a - b} = \dfrac{c + d}{c - d} \: \bigg \}}$

$\rm :\longmapsto\:\dfrac{sin(a + b)}{sin(a - b)} = \dfrac{k + 1}{k - 1}$

$\rm :\longmapsto\:(k - 1) \: sin(a + b) = (k + 1) \: sin(a - b)$

${{\boxed{\bf{Hence, Proved}}}}$

Additional Information :-

$\green{\boxed{\bf{sin(x + y) = sinxcosy + sinycosx}}}$

$\green{\boxed{\bf{sin(x - y) = sinxcosy - sinycosx}}}$

$\green{\boxed{\bf{cos(x - y) = cosxcosy + sinxsiny}}}$

$\green{\boxed{\bf{cos(x + y) = cosxcosy - sinxsiny}}}$

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

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

$\green{\boxed{\bf{cot(x + y) = \dfrac{cotx \: coty - 1}{coty + cotx} }}}$

$\green{\boxed{\bf{cot(x - y) = \dfrac{cotx \: coty + 1}{coty - cotx} }}}$