Question

If Sinxcosy=1/4 and 3tanx=4tany then sin(x-y)

Answer 1

Answer:

$sin \: ( \: x \: - \: y \: ) \: = \dfrac{1}{16}$

Step-by-step explanation:

Given---->

$sinx \: cosy \: = \dfrac{1}{4} \: and \: 3 \: tanx \: = 4 \: tany$

To find ---->

$value \: of \: sin \: ( \: x \: - \: y \: )$

Concept used ----->

1)

$tanx \: = \dfrac{sinx}{cosx}$

2)

$sin \: ( \: \alpha \: - \: \beta \: ) \: = \: sin \alpha \: cos \beta \: - \: cos \alpha \: sin \beta$

Solution----> ATQ ,

$\: \: \: \: \: \: 3 \: tanx \: = \: 4 \: tany \\ = >3 \: \dfrac{sinx}{cosx} \: = \: 4 \: \dfrac{siny}{cosy } \\ = > 3 \: sinx \: cosy \: = \: 4 \: cosx \: siny \\ putting \: sinx \: cosy \: = \dfrac{1}{4} \: in \: it \: we \: get \\ = > 3 \: ( \: \dfrac{1}{4} \: ) \: = \: 4 \: cosx \: siny \\ = > cosx \: siny \: = \dfrac{3}{4 \times 4} \\ = > cosx \: siny \: = \dfrac{3}{16} \\ now \: \\ sin ( \: x \: - \: y \: ) \: = \: sinx \: cosy \: - \: cosx \: siny \\ now \: putting \: sinx \: cosy \: = \dfrac{1}{4} \: and \: cosx \: siny \: = \dfrac{3}{16} \: we \: get \\ = > \: sin \: ( \: x \: - \: y \: ) \: = \: \dfrac{1}{4} \: - \frac{3}{16} \\ = > sin \: ( \: x \: - \: y \: ) \: = \dfrac{4 \: - \: 3}{16} \: \\ = > \: sin \: ( \: x \: - \: y \: ) \: = \dfrac{1}{16}$

Answer 2

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