If sin x-y-sin x-y-a-b-a-bthentan x-tan y-A- a-bB- b-aC- 1D- 0

The correct option is A sin(x+y)/sin(x y)=a+b/a b⇒sin(x+y)+sin(x y)/sin(x+y) sin(x y)=a+b+a b/a+b a+b=2a/2b=a/b⇒2sin x cos y/2 cos x sin y=a/b ⇒tan x cot y=a/ ...

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Byju's Answer

Standard XI

Mathematics

General Solution of tan theta = tan alpha

If sin x+y/si...

Question

If $\frac{s i n (x + y)}{s i n (x - y)} = \frac{a + b}{a - b} then \frac{t a n x}{t a n y} =$

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Solution

The correct option is A
$\frac{s i n (x + y)}{s i n (x - y)} = \frac{a + b}{a - b} \Rightarrow \frac{s i n (x + y) + s i n (x - y)}{s i n (x + y) - s i n (x - y)} = \frac{a + b + a - b}{a + b - a + b} = \frac{2 a}{2 b} = \frac{a}{b} \Rightarrow \frac{2 s i n x c o s y}{2 c o s x s i n y} = \frac{a}{b}$

$\Rightarrow tan x cot y = \frac{a}{b} \Rightarrow \frac{t a n x}{t a n y} = \frac{a}{b}$

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Similar questions

Q. If

\frac{s i n (x + y)}{s i n (x - y)} = \frac{a + b}{a - b}, then \frac{t a n x}{t a n y} =

Q. If

\frac{s i n (x + y)}{s i n (x - y)} = \frac{a + b}{a - b},

Show that

\frac{t a n x}{t a n y} = \frac{a}{b} .

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and

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Ifsin(x+y)sin(x−y)=a+ba−bthentan xtan y=

The correct option is A sin(x+y)sin(x−y)=a+ba−b⇒sin(x+y)+sin(x−y)sin(x+y)−sin(x−y)=a+b+a−ba+b−a+b=2a2b=ab⇒2sin x cos y2 cos x sin y=ab ⇒tan x cot y=ab⇒tan xtan y=ab

If $\frac{s i n (x + y)}{s i n (x - y)} = \frac{a + b}{a - b} then \frac{t a n x}{t a n y} =$