Question

secA + tana =1/2;then sinA=?

Answer 1

Given that,

$\longrightarrow\sf{\sec A+\tan A=\dfrac{1}{2}\quad\quad\dots(1)}$

We know that,

$\longrightarrow\sf{\sec^2A-\tan^2A=1}$

Factorising LHS,

$\longrightarrow\sf{(\sec A-\tan A)(\sec A+\tan A)=1}$

From (1),

$\longrightarrow\sf{\dfrac{1}{2}\cdot(\sec A-\tan A)=1}$

$\longrightarrow\sf{\sec A-\tan A=2\quad\quad\dots(2)}$

On adding (1) and (2), we get,

$\longrightarrow\sf{(\sec A+\tan A)+(\sec A-\tan A)=\dfrac{1}{2}+2}$

$\longrightarrow\sf{2\sec A=\dfrac{5}{2}}$

$\longrightarrow\sf{\sec A=\dfrac{5}{4}}$

$\longrightarrow\sf{\cos A=\dfrac{4}{5}\quad\quad\dots(3)}$

On subtracting (2) from (1), we get,

$\longrightarrow\sf{(\sec A+\tan A)-(\sec A-\tan A)=\dfrac{1}{2}-2}$

$\longrightarrow\sf{2\tan A=-\dfrac{3}{2}}$

$\longrightarrow\sf{\tan A=-\dfrac{3}{4}}$

$\longrightarrow\sf{\dfrac{\sin A}{\cos A}=-\dfrac{3}{4}}$

From (3),

$\longrightarrow\sf{\dfrac{\sin A}{\left(\dfrac{4}{5}\right)}=-\dfrac{3}{4}}$

$\longrightarrow\sf{\sin A=-\dfrac{3}{4}\cdot\dfrac{4}{5}}$

$\longrightarrow\sf{\underline{\underline{\sin A=-\dfrac{3}{5}}}}$