Question

if sinA =p/q find the value of tanA
​

Answer 1

Answer:

It should be:

TanA=1

Hope I am correct.

And this helps you.

Answer 2

Answer:

$\large\boxed{\sf{\tan A = \dfrac{p}{{q}^{2}-{p}^{2}}}}$

Step-by-step explanation:

Given

$\sin( \alpha ) = \frac{p}{q}$

To find $\tan \alpha$

We know that

$\cos( \alpha ) = \sqrt{1 - { \sin }^{2} \alpha }$

Therefore, we get

$= > \cos( \alpha ) = \sqrt{1 - {( \frac{p}{q} )}^{2} } \\ \\ = > \cos( \alpha ) = \sqrt{1 - \frac{ {p}^{2} }{ {q}^{2} } } \\ \\ = > \cos( \alpha ) = \sqrt{ \frac{ {q}^{2} - {p}^{2} }{ {q}^{2} } } \\ \\ = > \cos( \alpha ) = \frac{ \sqrt{ {q}^{2} - {p}^{2} } }{q}$

Now, we know that,

$\tan( \alpha ) = \dfrac{ \sin( \alpha ) }{ \cos( \alpha ) }$

Therefore, we will get

$= > \tan( \alpha ) = \dfrac{ \frac{p}{q} }{ \frac{ {q}^{2} - {p}^{2} }{q} } \\ \\ = > \tan( \alpha ) = \frac{p}{q} \times \frac{q}{ {q}^{2} - {p}^{2} } \\ \\ = > \tan( \alpha ) = \frac{p}{ {q}^{2} - {p}^{2} }$

Hence, $\bold\red{\tan A = \dfrac{p}{{q}^{2}-{p}^{2}}}$