Question

tanA²-sinA²=tanA²×sinA²
More questions coming from trigonometry. Thanks​

Answer 1

Step-by-step explanation:

Sin^2A/Cos^2A - Sin^2A

= Sin^2A(1-Cos^2A)/Cos^2A

Answer 2

QUESTION:

${ \tan( \alpha ) }^{2} - { \sin( \alpha ) }^{2} = { \tan( \alpha ) }^{2} \times { \sin( \alpha ) }^{2}$

I LET A AS ALPHA;

ANSWER:

WE use the trigonometry identity here;

$\tan( \alpha ) = \frac{ \sin( \alpha ) }{ \cos( \alpha ) }$

${ \sin( \alpha ) }^{2} + { \cos( \alpha ) }^{2} = 1$

now come to main question ;

TAKING LHS SIDE;

${ \tan( \alpha ) }^{2} - { \sin( \alpha ) }^{2}$

$\frac{ { \sin( \alpha ) }^{2} }{ { \cos( \alpha ) }^{2} } - { \sin( \alpha ) }^{2}$

TAKING lcm;

$\frac{ { \sin( \alpha ) }^{2} - { \sin( \alpha ) }^{2} \times { \cos( \alpha ) }^{2} }{ { \cos( \alpha ) }^{2} }$

taking common ;

$\frac{ { \sin( \alpha ) }^{2}(1 - { \cos( \alpha ) }^{2} }{ { \cos( \alpha ) }^{2} }$

$\frac{ { \sin( \alpha ) }^{2} (1 - { \cos( \alpha ) }^{2} }{ { \cos( \alpha ) }^{2} } \\ \\ \\ { \tan( \alpha ) }^{2} \times { \sin( \alpha ) }^{2}$

LHS = RHS

HENCE PROVED

PLZ MARK AS BRAINLIEST