If tanA -xtanB- Prove that sin A-B-sin A-B-x-1-x-1-

TanA=x tanB sinA/cosA= xsinB/cosB ⇒ sinAcosB=xcosAsinB Now, sin(A B)/sin(A+B)= sinAcosB sinBcosA/sinAcosB+cosAsinB = xcosAsinB cosA sinB/xcosAsinB+cosAsinB = co ...

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Byju's Answer

Standard XII

Mathematics

Sum of Trigonometric Ratios in Terms of Their Product

If tanA =xtan...

Question

If $t a n A = x t a n B,$ Prove that $\frac{s i n (A - B)}{s i n (A + B)} = \frac{x - 1}{x + 1}$ .

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Solution

$t a n A = x t a n B$
$\frac{s i n A}{c o s A} = \frac{x s i n B}{c o s B}$
$\Rightarrow s i n A c o s B = x c o s A s i n B$
Now,
$\frac{s i n (A - B)}{s i n (A + B)} = \frac{s i n A c o s B - s i n B c o s A}{s i n A c o s B + c o s A s i n B}$
$= \frac{x c o s A s i n B - c o s A s i n B}{x c o s A s i n B + c o s A s i n B}$
$= \frac{c o s A s i n B (x - 1)}{c o s A s i n B (x + 1)})$
$= \frac{x - 1}{x + 1}$
$∴ \frac{s i n (A - B)}{s i n (A + B)} = \frac{x - 1}{x + 1}$
Hence proved.

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If tanA=xtanB, Prove that sin(A−B)sin(A+B)=x−1x+1.

tanA=xtanB sinAcosA=xsinBcosB ⇒sinAcosB=xcosAsinB Now, sin(A−B)sin(A+B)=sinAcosB−sinBcosAsinAcosB+cosAsinB =xcosAsinB−cosAsinBxcosAsinB+cosAsinB =cosAsinB(x−1)cosAsinB(x+1)) =x−1x+1 ∴sin(A−B)sin(A+B)=x−1x+1 Hence proved.

If $t a n A = x t a n B,$ Prove that $\frac{s i n (A - B)}{s i n (A + B)} = \frac{x - 1}{x + 1}$ .