Question

(secA+cosA) (secA-cosA)=tanA*tanA+sinA*sinA

Answer 1

Hey Friends!!

Here is your answer↓⬇

⬇⬇⬇⬇⬇⬇⬇⬇⬇⬇⬇⬇⬇⬇


This question is solved by 2 methods:-)


$\huge \boxed{1st \: \: method:-)}$

$\bf = > ( \sec(x) + \cos(x) )( \sec(x) - \cos(x) ) = { \tan }^{2} x + { \sin }^{2} x.$


$\huge \bf \underline{Solving \: LHS}$

$\bf = { \sec }^{2} x - { \cos }^{2}x.$


$\bf = (1 + { \tan }^{2} x) - (1 - { \sin }^{2} x).$



$\bf = 1 + { \tan }^{2} x - 1 + { \sin }^{2} x.$


$\bf = { \tan }^{2} x + { \sin }^{2}x.$


$\huge \boxed{LHS = RHS.}$



$\huge \boxed{2nd \: \: method:-)}$


$\huge \bf \underline{Solving \: LHS.}$

$\bf = ( \sec(x) + \cos(x) )( \sec(x) - \cos(x) ) .$


$\bf = ( \frac{1}{ \cos(x) } + \cos(x) )( \frac{1}{ \cos(x) } - \cos(x) ).$



$\bf = ( \frac{1 + { \cos}^{2}x }{ \cos(x) } )( \frac{1 - { \cos }^{2} x}{ \cos(x) } ).$



$\bf = ( \frac{1 + { \cos }^{2} x}{ { \cos }^{2}x } )( { \sin }^{2}x).$



$\bf = ( \frac{1}{ { \cos }^{2}x } + \frac{ { \cos }^{2}x }{ { \cos }^{2}x } )( { \sin }^{2} x).$



$\bf = ( \frac{1}{ { \cos }^{2}x } + 1)( { \sin }^{2} x).$



$\bf = \frac{ { \sin }^{2} x}{ { \cos }^{2} x} + { \sin }^{2} x.$



$\bf = { \tan }^{2} x + { \sin }^{2} x.$


$\huge \boxed{LHS = RHS.}$

✅✅ Hence , it is proved ✔✔.



$\huge \boxed{THANKS}$



$\huge \bf \underline{Hope \: it \: is \: helpful \: for \: you}$