Question

If secA=$\frac{2}{\sqrt{3}}$, then prove that $\frac{tanA}{cosA}+\frac{1+sinA}{tanA}=\frac{4+9\sqrt{3}}{6}$

Answer 1

Given, secA = $\dfrac{2}{\sqrt{3}}$


secA = sec30°


A = 30°





Now,

$\dfrac{tanA}{cosA}=tanA \times \dfrac{1}{cosA} \\ \\\\ \dfrac{tanA}{cosA}=tanAsecA$


And, $\dfrac{1+sinA}{tanA}=\dfrac{1}{tanA}+\dfrac{sinA}{tanA}$


$\dfrac{1+sinA}{tanA}= cotA + \dfrac{sin}{\dfrac{sinA}{cosA}}$


$\dfrac{1+sinA}{tanA}= cotA + cosA$




Therefore,

$\frac{tanA}{cosA}+\frac{1+sinA}{tanA} = tanA secA + cotA + secA$



tanAsecA + cotA + cosA

tan30°$\dfrac{2}{\sqrt{3}} + cot30 \degree + \dfrac{\sqrt{3}}{2}$

$\dfrac{2}{\sqrt{3}} \dfrac{1}{\sqrt{3}} + \sqrt{3} + \dfrac{\sqrt{3}}{2}$


$\frac{2}{3} + \frac{2 \sqrt{3} + \sqrt{3} }{2} \\ \\ \\ \frac{4 + 9 \sqrt{3} }{6}$




Hence, proved .

Answer 2

sec A = 2/√3

secA = sec30°

A = 30°



Since the value of angle A is 30°,




tanA / cosA = tan30° / cos30° = ( 1/√3)/(√3/2) = 2 / 3



( 1 + sinA ) / tanA = ( 1 + sin30 )/tan30° = { 1 + ( 1 / 2 ) } / ( 1/√3) = { ( 3/2)/(1/√3) = 3√3 / 2



therefore,




$\frac{tanA}{cosA}+\frac{1+sinA}{tanA}= 2/3 + 3\sqrt{3} / 2$



$\frac{tanA}{cosA}+\frac{1+sinA}{tanA}=\dfrac{4+9\sqrt{3}}{6}$



$\textsf{ Proved. }$