Question

If sec A+tan A=x; then tan A:
(a)
$2 \: x$
(b)
$1 \: 2x$
(C) x² – 1 /2x
(d)
$2x \: x ^{2} - 1$
​

Answer 1

Given:-

$\red{➤}\:$$\sf \sec \:A+\tan\:A = x \:\_\:\_\:\_ eq(1)$

To Find:-

$\orange{☛}\:$$\sf Value\: of \; \tan \:A$

Solution:-

$\underline{\tt{Formula\:Applied}}$

$\green{ \underline { \boxed{ \sf{1+\tan² \theta = \sec²\theta}}}}$

$\implies\sf \sec²\theta - \tan²\theta =1$

$\\\underline{\tt\pink{Similarly-}}$

$\begin{gathered}\\\quad\longrightarrow\quad\sf \sec²\:A- \tan²\:A = 1 \\\end{gathered}$

$\begin{gathered}\\\quad\longrightarrow\quad\sf (\sec\:A- \tan\:A)(\sec\:A+\tan\:A) = 1 \quad(a²-b²= (a-b)(a+b))\\\end{gathered}$

$\begin{gathered}\\\quad\longrightarrow\quad\sf (\sec\:A- \tan\:A).x = 1 \quad(\sec\:A+\tan\:A=x)\\\end{gathered}$

$\begin{gathered}\\\quad\longrightarrow\quad\sf \sec\:A- \tan\:A = \frac{1}{x}\_\:\_\:\_ eq(2) \\\end{gathered}$

$\\\underline{\tt\red{Subtracting \:eq(2) \: from \:eq(1)-}}$

$\begin{gathered}\\\quad\longrightarrow\quad\sf (\sec\:A+\tan\:A)-(\sec\:A-\tan\:A)= x - \frac{1}{x} \\\end{gathered}$

$\begin{gathered}\\\quad\longrightarrow\quad\sf \cancel{\sec\:A}+\tan\:A\cancel{-\sec\:A}+\tan\:A = \frac{ x²-1}{x}\quad(Taking \;l.c.m. \: as \:x) \\\end{gathered}$

$\begin{gathered}\\\quad\longrightarrow\quad\sf 2\tan\:A= \frac{ x²-1}{x}\\\end{gathered}$

$\begin{gathered}\\\quad\longrightarrow\quad\boxed{\sf \tan\:A= \frac{ x²-1}{2x}}\\\\\end{gathered}$

$\therefore \sf{Option (c) \:\red{\dfrac{ x²-1}{2x}}\; is\:Correct }$✔️