Question

if sec A = 2/root3 then find the value of tanA/cosA+1+sinA/tanA

Answer 1

I hope it will help u.....

Answer 2

Answer:

$\frac{\tan A}{\cos A}+\frac{1+\sin A}{\tan A}=\frac{4+9\sqrt3}{6}$      

Step-by-step explanation:

Given : $\sec A=\frac{2}{\sqrt3}$

To find : $\frac{\tan A}{\cos A}+\frac{1+\sin A}{\tan A}$

Solution :

Using trigonometry identity,

$\sec A=\frac{2}{\sqrt3}=\frac{H}{B}$

Applying Pythagoras theorem,

$H^2=B^2+P^2$

$2^2=\sqrt{3}^2+P^2$

$4=3+P^2$

$P^2=1$

$P=1$

So, P=1 , H=2 , $B=\sqrt{3}$

We know, $\sin A=\frac{P}{H}$

$\sin A=\frac{1}{2}$

$\cos A=\frac{B}{H}$

$\cos A=\frac{\sqrt3}{2}$

$\tan A=\frac{P}{B}$

$\tan A=\frac{1}{\sqrt 3}$

Substitute all the values in the expression,

$=\frac{\frac{1}{\sqrt 3}}{\frac{\sqrt3}{2}}+\frac{1+\frac{1}{2}}{\frac{1}{\sqrt 3}}$

$=\frac{1\times 2}{\sqrt 3\times \sqrt3}+\frac{\frac{3}{2}}{\frac{1}{\sqrt 3}}$

$=\frac{1\times 2}{\sqrt 3\times \sqrt3}+\frac{3\times \sqrt3}{2\times 1}$

$=\frac{2}{3}+\frac{3\sqrt3}{2}$

$=\frac{4+9\sqrt3}{6}$

Therefore, $\frac{\tan A}{\cos A}+\frac{1+\sin A}{\tan A}=\frac{4+9\sqrt3}{6}$