If tan A - x -sin B -1- x cos B and tan B - y sin A -1- y cos A then sin A -sin B -A- x-y- y-x- x-yD- x-y

The correct option is A sin A/cos A = x sin B/1 x cos B⇒sin A x sin A cos B = x cos A sin B ⇒sin A = x sin (A+ B) Similarly sin B = y sin (A+B). ∴sin A/sin B ...

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Byju's Answer

Standard XI

Mathematics

Sin2a and Cos2a in Terms of Tana

If tan A = x ...

Question

If $tan A = \frac{x = sin B}{1 - x cos B}$ and $tan B = \frac{y sin A}{1 - y cos A}$ then $\frac{sin A}{sin B} =$

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Solution

The correct option is A
$\frac{sin A}{cos A} = \frac{x sin B}{1 - x cos B} \Rightarrow sin A - x sin A cos B = x cos A sin B \Rightarrow sin A = x sin (A + B)$
Similarly $sin B = y sin (A + B) . ∴ \frac{sin A}{sin B} = \frac{x}{y} .$

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If tanA=x=sinB1−xcosB and tanB=ysinA1−ycosA then sinAsinB=

The correct option is A sinAcosA=xsinB1−xcosB⇒sinA−xsinAcosB=xcosAsinB⇒sinA=xsin(A+B) Similarly sinB=ysin(A+B).∴sinAsinB=xy.

If $tan A = \frac{x = sin B}{1 - x cos B}$ and $tan B = \frac{y sin A}{1 - y cos A}$ then $\frac{sin A}{sin B} =$

The correct option is A
$\frac{sin A}{cos A} = \frac{x sin B}{1 - x cos B} \Rightarrow sin A - x sin A cos B = x cos A sin B \Rightarrow sin A = x sin (A + B)$
Similarly $sin B = y sin (A + B) . ∴ \frac{sin A}{sin B} = \frac{x}{y} .$