If tan-a sin-1-a cos- and tan-b sin-1-b cos- then show that sin-sin-ab-

We have, #10; #10; tanα =asinβ1 acosβ #10; #10; sinαcosα=asinβ #10;1 acosβ #10; #10; sinα asinαcosβ =acosαsin #10;β #160; #1 ...

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Basic Inverse Trigonometric Functions

If tanα=a s...

Question

If $tan α = \frac{a sin β}{1 - a cos β}$ and $tan β = \frac{b sin α}{1 - b cos α}$ then show that $\frac{sin α}{sin β} = \frac{a}{b}$ .

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tan α = \frac{a sin β}{1 - a cos β} a n d tan β = \frac{b sin α}{1 - b cos α}

\frac{sin α}{sin β} = \frac{a}{b}

I f tan α = \frac{a sin β}{1 - a cos β} a n d tan β = \frac{b sin α}{1 - b cos α}

\frac{sin α}{sin β} = \frac{a}{b .}

(b cos α, b sin α)

(a cos β, a sin β)

M (x, y)

A M : M B = b : a

x cos \frac{α + β}{2} + y sin \frac{α + β}{2} =

tan β = \frac{2 sin α sin γ}{sin (α + γ)}

tan α, t a n β, t a n γ

H . P

A

(b cos α, b sin α)

B

(a cos β, a sin β)

M

(x, y)

A M

B M

b : a

x + y tan (\frac{α + β}{2}) = 0

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