Question

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


`1 - sin {}^(2) x * 1 + \tan( { }^(2) ) x = 1`
Prove that :
Please help i need the right one​

Answer 1

`\tt (1 - sin ^2x)*(1+tan^2x) = 1`

L.H.S

`\tt cos^2x * sec^2x`

`\tt cos^2x * (1)/(cos^2x)`

`\tt \cancel{ cos^2x} *\cancel{ (1)/(cos^2x)}`

`\tt 1`

R.H.S

Hence proved

Answer 2

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.