Question

SinA÷ sinB=p and cosA÷cosB=q,. Then find tanA

Answer 1

Answer:

Step-by-step explanation:

SinA/sinB=p

SinA=psinB

cosA/cosB=q

CosA=qcosB

TanA=sinA/cosA

TanA=psinB/qcosB

TanA=tanB.p/q

Answer 2

Hi!

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Given :

$\frac{SinA}{SinB} = P$

SinA = P SinB -----> 1

$\frac{CosA}{CosB} = q$

CosA = q CosB -----> 2

Dividing Equation (1) and (2)

$\frac{SinA}{CosA} = \frac{P}{q} \frac{SinB}{CosB}$

$tanA = \frac{P}{q} * tanB$

$\frac{tanA}{P} = \frac{tanB}{q} = K$

tanA = PK , tanB = qK

From eq (1)

SinA = P SinB

Squaring Both Sides

$Sin^2A = P^2 Sin^2B$

$\frac{1 -Cos2 A}{2} = P^2\ \frac{1 - Cos2B}{2}$

=> $( 1 - Cos2A ) = P^2 (1 - Cos2B )$

$Cos2\theta = \frac{1 - tan^2 \theta }{1 + tan^2 \theta`}$

=> $(1 - \frac{1- - tan^2A}{1 + tan^2A} ) = P^2 (1 - \frac{1 - tan^2B}{1 + tan^2B} )$

=> $( \frac{2tan^2A}{1 + tan^2 A} )= P^2 ( \frac{2 tan^2B}{1 + tan^2B} )$

=> $\frac{2 * P^2 K^2 }{1 + P^2 K^2 }$ = $P^2 ( \frac{2q^2 k^2 }{1 + q^2k^2} )$

=> $\frac{1}{1 + p^2k^2} = \frac{q^2}{1 + q^2k^2}$

=> $1 + q^2 k^2 = q^2 + p^2q^2k^2$

=> $1 - q^2 = p^2q^2k^2 - q^2k^2$

=> $1 - q^2 = q^2k^2 ( p^2 -1 )$

=> $k^2 = \frac{1}{q^2} \ ( \frac{1 - q^2 }{p^2 - 1 } )$

=> $k = \frac{1}{q} \ \sqrt{\frac{1 - q^2}{p^2 - 1 } }$

=> tanA = pk => $\frac{P}{q} \ \sqrt{\frac{1 -q^2 }{p^2- 1} }$

⇒ tanB = qk => $\frac{q}{q} \ \sqrt{\frac{1 -q^2 }{p^2 - 1 } }$

q\q get Cancel out

We have ,

=> $tanB =\sqrt{\frac{1 -q^2 }{p^2 - 1 }$

=> tanA = pk => $\frac{P}{q} \ \sqrt{\frac{1 -q^2 }{p^2- 1} }$

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Thanks !!