Question

$\huge\color{pink}\mathfrak{Help\:Needed}$
$\sf\color{red} {Prove\:That\: {sec}^{2} - {cos}^{2} = {tan}^{2} + {sin}^{2} }$


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Answer 1

Step-by-step explanation:

Here is your answer ☺️

If the LHS of the given equation is: sec2a-sin2atan2a =sec 2 atan -sin2atan2a =1cos2a. cos2asin2a-sin2a.cos2asin2a =1si 1sin2a-cos2a-cosec2a-cos2a

Answer 2

$\huge\rm\underline\purple{⟹ᴀɴsᴡᴇʀ :-}$

$\frac{{ \cos( \alpha )}^{2}+{ \tan( \alpha ) }^{2} - 1}{ \ \sin( \ { \alpha }^{2} ) } sin( α2 )cos(α) 2 +tan(α) 2 −1$

Putting sin^2 theta+cos^2 theta instead of 1

$→\frac{ { \cos( \alpha ) }^{2} + { \tan( \alpha ) }^{2} - { \cos( \alpha ) }^{2} - { \sin( \alpha ) }^{2} }{ { \sin( \alpha ) }^{2} }sin(α) 2cos(α)2+tan(α−cos(α)2 −sin(α) 2$

tan^2 theta - sin^2 theta/ sin^2 theta

$→\frac{ { \tan( \alpha ) }^{2} }{ { \sin( \alpha ) }^{2} } - \frac{ { \sin( \alpha ) }^{2} }{ { \sin( \alpha ) }^{2} }sin(α) 2 tan(α) 2− sin(α) 2 sin(α)2$

$→\frac{ { \sin( \alpha ) }^{2} }{ { \cos( \alpha ) }^{2} + { \sin( \alpha ) }^{2} } - 1 cos(α) 2 +sin(α) 2 sin(α) 2−1$

$→\frac{1}{ { \cos( \alpha ) }^{2} } - 1 cos(α) 2 1 −1$

${ \sec( \alpha ) }^{2} - 1 = { \tan( \alpha ) }^{2}sec(α)2−1=tan(α) 2$

Hence PROVED!!!