Question

If secA= 2 then find the value of tanA +cotA is

Answer 1

Solution:-

$\rm \implies \sec A = \dfrac{2}{1} = \dfrac{h}{b}$

We have

$\implies \rm \: h = 2,b = 1 \: and \: p = x$

We have to find Perpendicular (p), we use pythagoras theorem

$\rm \implies \: {h}^{2} = {p}^{2} + {b}^{2}$

$\rm \implies \: (2) {}^{2} = {p}^{2} + (1) ^{2}$

$\rm \implies \: 4 = {p}^{2} + 1$

$\rm \implies \: {p}^{2} = 4 - 1 = 3$

$\rm \implies \: p = \sqrt{ 3}$

now we get

$\implies \rm \: h = 2,b = 1 \: and \: p = \sqrt{3}$

We have to find

$\rm \implies \tan A + \cot A$

we know that

$\rm \implies \tan A = \dfrac{p}{b}$

$\rm \implies \: \cot A = \dfrac{b}{p}$

So we can write as

$\rm \implies \: \dfrac{p}{b} + \dfrac{b}{p}$

$\rm \implies \: \dfrac{ {p}^{2} + {b}^{2} }{bp}$

Put the value of p and b

$\rm \implies \dfrac{( { \sqrt{3} )}^{2} + 1 }{ \sqrt{3} }$

$\rm \implies \: \dfrac{3 + 1}{3}$

$\rm \implies \: \dfrac{4}{3}$

Answer

$\rm \implies \: \dfrac{4}{3}$