Question

-tan^2x+sec^2x= 1 ?
match the right to the left. help ?

Answer 1

\[- \tan ^{2}x+\sec ^{2}x=-y ^{2}+\left( \sqrt{1+y ^{2}} \right)^{2}=-y ^{2}+1+y ^{2}=1\]

Answer 2

thank you

Answer 3

@surjithayer

Answer 4

leme make this a breeze for you,

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

we elaborate this further by expanding tan and sec

\[-\frac{ \sin ^{2}x }{ \cos ^{2}x }+\frac{ 1 }{ \cos ^{2}x }=1\]
we group the numerators above the already common denominator cos^2x
\[\frac{ 1-\sin ^{2}x }{ \cos ^{2}x }=1\]
we switch the numerators to get a better view of an already known trigonometric identity (1-sin^2x=cos^2x)
\[\frac{ \cos ^{2}x }{ \cos ^{2}x }=1\]
we divide cos^2x by itself, wich of course gives 1
\[1=1\]