Question

(1-sin^2x) (1+tan^2x)=1

Answer 1

$\tt (1 - sin ^2x)\times(1+tan^2x) = 1$

L.H.S

$\tt cos^2x \times sec^2x$

$\tt cos^2x \times \frac{1}{cos^2x}$

$\tt \cancel{ cos^2x} \times\cancel{ \frac{1}{cos^2x}}$

$\tt 1$

R.H.S

Hence proved

Answer 2

Step-by-step explanation:

${ \tt{ \blue{(1 - { \sin}^{2}x )(1 + { \tan }^{2}x) = 1 }}}$

• Remember → 1 + tan²x = sec²x

$= { \tt{ \blue{(1 - { \sin }^{2}x)( { \sec}^{2}x) }}}$

• But sec²x → 1/cos²x

$= { \tt{ \blue{(1 - { \sin}^{2} x)( \frac{1}{ \cos {}^{2} x} )}} }\\ \\ = { \blue {\tt{ \frac{1 - { \sin }^{2}x }{ { \cos }^{2}x } }}}$

• Remember from the first identity of trignometry;

[ cos²x + sin²x = 1 ].

• Therefore: [ cos²x = 1 - sin²x ]

$= { \blue{\rm{ \frac{1 - { \sin}^{2}x }{1 - { \sin }^{2}x } }}} \\ \\ { \rm{ \blue{= 1 \: }}}$

Hence proved.