Question

If sin A = √3, then tan A is _,​

Answer 1

Given sin A = √3

So we can write sinA as

$sin \: a = \frac{perpendicular \: }{hypotenus}$

So in the attachment given A right triangle ∆ PQA

with angle R= theta

Now,

$= > sin \: a = \frac{perpendicular}{hypotenus}$

$= > sin \: a \: = \frac{pq}{pa}$

$= > \: sin \:a = \sqrt{3} \\ \\ = > \: \frac{pq}{pa} = \sqrt{3} \\ \\ = > pq = \sqrt{3} and \: qa = 1$

So now by applying Pythagoras formula we can find the base of the ∆✓

Which is as follows:

.

$= > qa = \sqrt{pq ^{2} - pa ^{2} }$

$= > qa = \sqrt{ (\sqrt{3} ) ^{2} - 1 ^{2} }$

$= > qa = \sqrt{3 - 1} \\ \\ = > qa = \sqrt{2}$

So we know that

.

$tan \: a = \frac{perpendicular}{base}$

$= > tan \: a = \frac{ \sqrt{3} }{ \sqrt{2} } \\ \\ = > tan \: a = \frac{ \sqrt{3} \times \sqrt{3} }{ \sqrt{2} \times \sqrt{2} } ......(rationalising) \\ \\ = > tan \: a = \frac{6}{2} = 3$

Therefore the value of tanA= 3

.

.

.

Answer 2

sin theta + cos theta = √3

( sin theta + cos theta )^ 2 = (√3 ) ^2

sin^2 theta + cos^2 theta +2sin theta. cos theta = 3

1 + 2 sin theta . cos theta = 3

2 sin theta . cos theta = 2

sin theta . cos theta = 1

sin theta . cos theta = sin^2 theta + cos^2 theta

( 1 = sin^2 theta + cos^2 theta

sin theta . cos theta / sin theta . cos theta

= sin^2 theta + cos^2 theta / sin theta . cos theta

1 = tan theta + cot theta