Question

if a sinθ+b cosθ= C then a-b tanθ/b+a tan θ =?

Answer 1

Answer:

$\large \text{ \dfrac{a-b\tan\theta}{b+a\tan\theta}=\dfrac{\pm \sqrt{a^2+b^2-c^2}}{c} }$

Step-by-step explanation:

Given

a sin θ  + b cos θ  = c

Let     a cos θ  -  b sin θ  = k

Now squaring on both sides and add both

$\large a^2\sin^2 \theta + b^2\cos^2 \theta +b^2\sin^2 \theta + a^2\cos^2 \theta=c^2+k^2\\\\\\\large a^2+b^2=k^2+c^2\\\\\\\large k=\pm\sqrt{a^2+b^2-c^2}$

Now

$\large \dfrac{a-b\tan\theta}{b+a\tan\theta}=\dfrac{a \cos \theta +b \sin\theta}{a \sin \theta +b \cos\theta}=\dfrac{k}{c}\\\\\\\large \text{Put the value here}\\\\\\\large \text{ \dfrac{a-b\tan\theta}{b+a\tan\theta}=\dfrac{\pm \sqrt{a^2+b^2-c^2}}{c} }$

Thus we get answer.

Answer 2

$\boxed{Explained\:Answer}$

______________________________

tan θ = a/b

sinθ/cosθ = a/b------(1)

lhs = (a sinθ- b cos θ)/(a sinθ + b cosθ)

= (asinθ/cosθ - bcosθ/cosθ)/(asinθ/cosθ+ bcosθ/cosθ)

dividing each term with cosθ

=(asinθ/cosθ-b/1)/(asinθ/cosθ+b/1)

=(a*a/b-b/1)/(a*a/b +b/1)

[from (1)]

= (a^2-b^2)/(a^2+b^2)

=RHS