Question

tan⁻¹(acosx-bsinx/bcosx+asinx)=tan⁻¹(a/b) - x, - π/2 < x < π/2,a/b tanx > -1,Prove it

Answer 1

we have to prove that,

tan⁻¹(acosx-bsinx/bcosx+asinx)=tan⁻¹(a/b) - x, where -π/2 < x < π/2 , a/b tanx > -1

Let tan⁻¹(a/b) - x = A .........(1)

⇒tan⁻¹(a/b) = x + A

⇒a/b = tan(x + A)

⇒a/b = (tanx + tanA)/(1 - tanx.tanA)

⇒a - atanx.tanA = btanx + btanA

⇒a - btanx = btanA + atanx.tanA

⇒(a - btanx) = tanA(b + atanx)

⇒(a - btanx)/(b + atanx) = tanA

putting, tanx = sinx/cosx

⇒(acosx - bsinx)/(bcosx + asinx) = tanA

⇒tan⁻¹ [(acosx - bsinx)/(bcosx + asinx)] = A ........(2)

from equations (1) and (2),

tan⁻¹[(acosx-bsinx)/(bcosx+asinx)]=tan⁻¹(a/b) - x [ proved]

Answer 2

Answer:

${tan}^{ - 1} ( \frac{acosx - bsinx}{bcosx + asinx} ) \\ \\ = {tan}^{ - 1} ( \frac{ \frac{acosx - bsinx}{bcosx} }{ \frac{bcosx + asinx}{bcosx} } ) \\ \\ = {tan}^{ - 1} ( \frac{ \frac{a}{b} - tanx }{1 + \frac{a}{b}tanx } ) \\ \\ = {tan}^{ - 1} \frac{a}{b} - {tan}^{ - 1} tanx \\ \\ = {tan}^{ - 1} \frac{a}{b} - x$