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(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​

2 Answers

2 votes


\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

User Thomas Liu
by
8.5k points
1 vote

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.

User Rocky Li
by
7.9k points

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