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(1-sin^2x) (1+tan^2x)=1
[tex]1 - sin {}^{2} x \times 1 + \tan( { }^{2} ) x = 1[/tex]
Prove that :
Please help i need the right one​

Respuesta :

dansontrnn

[tex]\tt (1 - sin ^2x)\times(1+tan^2x) = 1[/tex]

L.H.S

[tex]\tt cos^2x \times sec^2x[/tex]

[tex]\tt cos^2x \times \frac{1}{cos^2x}[/tex]

[tex]\tt \cancel{ cos^2x} \times\cancel{ \frac{1}{cos^2x}}[/tex]

[tex]\tt 1[/tex]

R.H.S

Hence proved

Step-by-step explanation:

[tex]{ \tt{ \blue{(1 - { \sin}^{2}x )(1 + { \tan }^{2}x) = 1 }}}[/tex]

• Remember → 1 + tan²x = sec²x

[tex] = { \tt{ \blue{(1 - { \sin }^{2}x)( { \sec}^{2}x) }}}[/tex]

• But sec²x → 1/cos²x

[tex] = { \tt{ \blue{(1 - { \sin}^{2} x)( \frac{1}{ \cos {}^{2} x} )}} }\\ \\ = { \blue {\tt{ \frac{1 - { \sin }^{2}x }{ { \cos }^{2}x } }}}[/tex]

• Remember from the first identity of trignometry;

[ cos²x + sin²x = 1 ].

• Therefore: [ cos²x = 1 - sin²x ]

[tex] = { \blue{\rm{ \frac{1 - { \sin}^{2}x }{1 - { \sin }^{2}x } }}} \\ \\ { \rm{ \blue{= 1 \: }}}[/tex]

Hence proved.

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