Question

If sec A+tanA=x then find the value of sinA

Answer 1

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

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On adding (1) and (2),

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$\sec A=\dfrac {\left (x+\dfrac {1}{x}\right)}{2}\\\\\\\sec A=\dfrac {x^2+1}{2x}$

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And on subtracting (2) from (1),

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$\tan A=\dfrac {\left (x-\dfrac {1}{x}\right)}{2}\\\\\\\sin A\sec A=\dfrac {x^2-1}{2x}\\\\\\\dfrac {\sin A(x^2+1)}{2x}=\dfrac {x^2-1}{2x}\\\\\\\large\text { \underline {\underline {\sin A=\dfrac {x^2-1}{x^2+1}}} }$