Proving the identity $(tan^2(x)+1)(cos^2(-x)-1)=-tan^2(x)$Trigonometric equality: $frac1 + sin A - cos A1 + sin A + cos A = tan fracA2$Prove $fracsin(A+B)cos(A-B)=fractan A+tan B1+tan Atan B$Prove trigonometry identity for $sin A+cos A$Proving $sin^2x+cos^2x=1$Proving tan((x + y)/2) = (sin x + sin y)/(cos x + cos y) with the angle sum and difference identitiesProve Trig IdentitiesHow to simplify a trigonometric expression with the identitiesDetermine exact values using angle sum and difference identity (trig)Trigonometry Exact Value using Half Angle IdentityDoes the trigonometric identity $cos^2(theta)+sin^2(theta)=1$ apply even when $theta$ is not in radians or degrees but simply a fraction?
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Proving the identity $(tan^2(x)+1)(cos^2(-x)-1)=-tan^2(x)$
Trigonometric equality: $frac1 + sin A - cos A1 + sin A + cos A = tan fracA2$Prove $fracsin(A+B)cos(A-B)=fractan A+tan B1+tan Atan B$Prove trigonometry identity for $sin A+cos A$Proving $sin^2x+cos^2x=1$Proving tan((x + y)/2) = (sin x + sin y)/(cos x + cos y) with the angle sum and difference identitiesProve Trig IdentitiesHow to simplify a trigonometric expression with the identitiesDetermine exact values using angle sum and difference identity (trig)Trigonometry Exact Value using Half Angle IdentityDoes the trigonometric identity $cos^2(theta)+sin^2(theta)=1$ apply even when $theta$ is not in radians or degrees but simply a fraction?
$begingroup$
Proving the trigonometric identity $(tan^2x+1)(cos^2(-x)-1)=-tan^2x$ has been quite the challenge. I have so far attempted using simply the basic trigonometric identities based on the Pythagorean Theorem. I am unsure if these basic identities are unsuitable for the situation or if I am not looking at the right angle to tackle this problem.
trigonometry
edited Mar 22 at 8:31
Eevee Trainer
10.1k31742
asked Mar 22 at 8:23
JamesJames
425
$endgroup$
add a comment |
$begingroup$
Proving the trigonometric identity $(tan^2x+1)(cos^2(-x)-1)=-tan^2x$ has been quite the challenge. I have so far attempted using simply the basic trigonometric identities based on the Pythagorean Theorem. I am unsure if these basic identities are unsuitable for the situation or if I am not looking at the right angle to tackle this problem.
trigonometry
edited Mar 22 at 8:31
Eevee Trainer
10.1k31742
asked Mar 22 at 8:23
JamesJames
425
$endgroup$
2
$begingroup$
$tan^2(x)+1=sec^2(x)=frac 1 cos^2(x)$ and $1-cos^2(-x)=1-cos^2(x)=sin^2(x)$.
$endgroup$
– Kavi Rama Murthy
Mar 22 at 8:25
1
$begingroup$
I wonder how you failed to solve this using the Pythagorean theorem, this is straightforward.
$endgroup$
– Yves Daoust
Mar 22 at 8:32
add a comment |
$begingroup$
Proving the trigonometric identity $(tan^2x+1)(cos^2(-x)-1)=-tan^2x$ has been quite the challenge. I have so far attempted using simply the basic trigonometric identities based on the Pythagorean Theorem. I am unsure if these basic identities are unsuitable for the situation or if I am not looking at the right angle to tackle this problem.
trigonometry
edited Mar 22 at 8:31
Eevee Trainer
10.1k31742
asked Mar 22 at 8:23
JamesJames
425
$endgroup$
Proving the trigonometric identity $(tan^2x+1)(cos^2(-x)-1)=-tan^2x$ has been quite the challenge. I have so far attempted using simply the basic trigonometric identities based on the Pythagorean Theorem. I am unsure if these basic identities are unsuitable for the situation or if I am not looking at the right angle to tackle this problem.
trigonometry
trigonometry
edited Mar 22 at 8:31
Eevee Trainer
10.1k31742
asked Mar 22 at 8:23
JamesJames
425
edited Mar 22 at 8:31
Eevee Trainer
10.1k31742
asked Mar 22 at 8:23
JamesJames
425
edited Mar 22 at 8:31
Eevee Trainer
10.1k31742
edited Mar 22 at 8:31
Eevee Trainer
10.1k31742
edited Mar 22 at 8:31
Eevee Trainer
10.1k31742
10.1k31742
asked Mar 22 at 8:23
JamesJames
425
asked Mar 22 at 8:23
JamesJames
425
asked Mar 22 at 8:23
JamesJames
425
425
2
$begingroup$
$tan^2(x)+1=sec^2(x)=frac 1 cos^2(x)$ and $1-cos^2(-x)=1-cos^2(x)=sin^2(x)$.
$endgroup$
– Kavi Rama Murthy
Mar 22 at 8:25
1
$begingroup$
I wonder how you failed to solve this using the Pythagorean theorem, this is straightforward.
$endgroup$
– Yves Daoust
Mar 22 at 8:32
add a comment |
2
$begingroup$
$tan^2(x)+1=sec^2(x)=frac 1 cos^2(x)$ and $1-cos^2(-x)=1-cos^2(x)=sin^2(x)$.
$endgroup$
– Kavi Rama Murthy
Mar 22 at 8:25
1
$begingroup$
I wonder how you failed to solve this using the Pythagorean theorem, this is straightforward.
$endgroup$
– Yves Daoust
Mar 22 at 8:32
2
2
$begingroup$
$tan^2(x)+1=sec^2(x)=frac 1 cos^2(x)$ and $1-cos^2(-x)=1-cos^2(x)=sin^2(x)$.
$endgroup$
– Kavi Rama Murthy
Mar 22 at 8:25
$begingroup$
$tan^2(x)+1=sec^2(x)=frac 1 cos^2(x)$ and $1-cos^2(-x)=1-cos^2(x)=sin^2(x)$.
$endgroup$
– Kavi Rama Murthy
Mar 22 at 8:25
1
1
$begingroup$
I wonder how you failed to solve this using the Pythagorean theorem, this is straightforward.
$endgroup$
– Yves Daoust
Mar 22 at 8:32
$begingroup$
I wonder how you failed to solve this using the Pythagorean theorem, this is straightforward.
$endgroup$
– Yves Daoust
Mar 22 at 8:32
add a comment |
5 Answers
5
active
oldest
votes
$begingroup$
Hint:
Multiply the two members by $cos^2x$ (certainly nonzero):
$$(sin^2x+cos^2x)(cos^2x-1)=-sin^2x.$$
answered Mar 22 at 8:34
Yves DaoustYves Daoust
132k676230
$endgroup$
$begingroup$
I do not believe your work is correct.
$endgroup$
– James
Mar 22 at 8:41
3
$begingroup$
@James I strongly believe his work is correct and elegant.
$endgroup$
– Le Anh Dung
Mar 22 at 8:43
1
$begingroup$
@James: my approach is fully correct. Sorry to say, you don't understand the principle of proving identities. You can rewrite keeping the denominators, but this is unnecessary. By the way, we can multiply by $cos^2x$ as this factor is guaranteed to be nonzero.
$endgroup$
– Yves Daoust
Mar 22 at 8:48
2
$begingroup$
@James: It is perfectly acceptable to prove an identity by transforming it in a reversible way. With this case, in multiplying-through by a non-zero expression ($cos^2x$), Yves reduced the identity to an utter triviality; to recapture the original identity, one could simply reverse the process and divide the triviality by the same non-zero expression ... even so, the essence of the identity is already established in the transformed version. (Note: We may assume $cos^2 x$ is non-zero, since the appearance of $tan x$ restricts the domain of consideration appropriately.)
$endgroup$
– Blue
Mar 22 at 9:28
1
$begingroup$
@James: if $cne0$, $a=biff ca=cb$. This is what you are missing. You don't have the correct experience in proving identities (no offense but I urge you to review your understanding of equivalence of identities), and you disregarded my comment that explained how to reshape the proof to your taste.
$endgroup$
– Yves Daoust
Mar 22 at 9:34
|
show 7 more comments
$begingroup$
Expanding gives $sin^2x-tan^2x+cos^2x-1$. The undesired terms cancel because $sin^2x+cos^2x=1$.
answered Mar 22 at 8:29
J.G.J.G.
32.9k23250
$endgroup$
add a comment |
$begingroup$
Recall:
$$beginalign
tan^2(x) - sec^2(x) &= -1 \
sin^2(x) + cos^2(x) &= 1
endalign$$
Thus,
$$beginalign
tan^2(x) + 1 &= sec^2(x)\
cos^2(x) - 1 &= -sin^2(x)
endalign$$
We also note that $cos(x)$ is an even function, and thus $cos(-x) = cos(x)$. Thus, the formula becomes:
$$(tan^2(x) + 1)(cos^2(-x) - 1) = -sec^2(x)sin^2(x) = - fracsin^2(x)cos^2(x) = -tan^2(x)$$
answered Mar 22 at 8:30
Eevee TrainerEevee Trainer
10.1k31742
$endgroup$
add a comment |
$begingroup$
- $1+tan^2x=sec^2x$
- $sin^2x+cos^2x=1$
$cos x=cos(-x)$ i.e. $cos x$ is an even function.- $sec x=1/cos x$
$$underbraceleft(1+tan^2xright)_=sec^2xunderbraceleft(cos^2(-x)-1right)_cos xtext is even function=-sec^2xcdotsin^2x=-tan^2x $$
answered Mar 22 at 8:31
Paras KhoslaParas Khosla
2,883523
$endgroup$
add a comment |
$begingroup$
We know
$(tan x)' = 1 + tan^2 x = frac1cos^2 x$ and- $cos (-x) = cos x$
Now, it follows immediately
begineqnarray* (tan^2(x)+1)(cos^2(-x)-1)
& = & (tan x)'left(frac1(tan x)'-1right) \
& = & 1 - (tan x)' \
& = & 1 - (1 + tan ^2 x) \
& = & - tan ^2 x \
endeqnarray*
answered Mar 22 at 9:11
trancelocationtrancelocation
13.6k1829
$endgroup$
add a comment |
Your Answer
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5 Answers
5
active
oldest
votes
5 Answers
5
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Hint:
Multiply the two members by $cos^2x$ (certainly nonzero):
$$(sin^2x+cos^2x)(cos^2x-1)=-sin^2x.$$
answered Mar 22 at 8:34
Yves DaoustYves Daoust
132k676230
$endgroup$
$begingroup$
I do not believe your work is correct.
$endgroup$
– James
Mar 22 at 8:41
3
$begingroup$
@James I strongly believe his work is correct and elegant.
$endgroup$
– Le Anh Dung
Mar 22 at 8:43
1
$begingroup$
@James: my approach is fully correct. Sorry to say, you don't understand the principle of proving identities. You can rewrite keeping the denominators, but this is unnecessary. By the way, we can multiply by $cos^2x$ as this factor is guaranteed to be nonzero.
$endgroup$
– Yves Daoust
Mar 22 at 8:48
2
$begingroup$
@James: It is perfectly acceptable to prove an identity by transforming it in a reversible way. With this case, in multiplying-through by a non-zero expression ($cos^2x$), Yves reduced the identity to an utter triviality; to recapture the original identity, one could simply reverse the process and divide the triviality by the same non-zero expression ... even so, the essence of the identity is already established in the transformed version. (Note: We may assume $cos^2 x$ is non-zero, since the appearance of $tan x$ restricts the domain of consideration appropriately.)
$endgroup$
– Blue
Mar 22 at 9:28
1
$begingroup$
@James: if $cne0$, $a=biff ca=cb$. This is what you are missing. You don't have the correct experience in proving identities (no offense but I urge you to review your understanding of equivalence of identities), and you disregarded my comment that explained how to reshape the proof to your taste.
$endgroup$
– Yves Daoust
Mar 22 at 9:34
|
show 7 more comments
$begingroup$
Hint:
Multiply the two members by $cos^2x$ (certainly nonzero):
$$(sin^2x+cos^2x)(cos^2x-1)=-sin^2x.$$
answered Mar 22 at 8:34
Yves DaoustYves Daoust
132k676230
$endgroup$
$begingroup$
I do not believe your work is correct.
$endgroup$
– James
Mar 22 at 8:41
3
$begingroup$
@James I strongly believe his work is correct and elegant.
$endgroup$
– Le Anh Dung
Mar 22 at 8:43
1
$begingroup$
@James: my approach is fully correct. Sorry to say, you don't understand the principle of proving identities. You can rewrite keeping the denominators, but this is unnecessary. By the way, we can multiply by $cos^2x$ as this factor is guaranteed to be nonzero.
$endgroup$
– Yves Daoust
Mar 22 at 8:48
2
$begingroup$
@James: It is perfectly acceptable to prove an identity by transforming it in a reversible way. With this case, in multiplying-through by a non-zero expression ($cos^2x$), Yves reduced the identity to an utter triviality; to recapture the original identity, one could simply reverse the process and divide the triviality by the same non-zero expression ... even so, the essence of the identity is already established in the transformed version. (Note: We may assume $cos^2 x$ is non-zero, since the appearance of $tan x$ restricts the domain of consideration appropriately.)
$endgroup$
– Blue
Mar 22 at 9:28
1
$begingroup$
@James: if $cne0$, $a=biff ca=cb$. This is what you are missing. You don't have the correct experience in proving identities (no offense but I urge you to review your understanding of equivalence of identities), and you disregarded my comment that explained how to reshape the proof to your taste.
$endgroup$
– Yves Daoust
Mar 22 at 9:34
|
show 7 more comments
$begingroup$
Hint:
Multiply the two members by $cos^2x$ (certainly nonzero):
$$(sin^2x+cos^2x)(cos^2x-1)=-sin^2x.$$
answered Mar 22 at 8:34
Yves DaoustYves Daoust
132k676230
$endgroup$
Hint:
Multiply the two members by $cos^2x$ (certainly nonzero):
$$(sin^2x+cos^2x)(cos^2x-1)=-sin^2x.$$
answered Mar 22 at 8:34
Yves DaoustYves Daoust
132k676230
edited Mar 22 at 10:26
answered Mar 22 at 8:34
Yves DaoustYves Daoust
132k676230
answered Mar 22 at 8:34
Yves DaoustYves Daoust
132k676230
answered Mar 22 at 8:34
Yves DaoustYves Daoust
132k676230
132k676230
$begingroup$
I do not believe your work is correct.
$endgroup$
– James
Mar 22 at 8:41
3
$begingroup$
@James I strongly believe his work is correct and elegant.
$endgroup$
– Le Anh Dung
Mar 22 at 8:43
1
$begingroup$
@James: my approach is fully correct. Sorry to say, you don't understand the principle of proving identities. You can rewrite keeping the denominators, but this is unnecessary. By the way, we can multiply by $cos^2x$ as this factor is guaranteed to be nonzero.
$endgroup$
– Yves Daoust
Mar 22 at 8:48
2
$begingroup$
@James: It is perfectly acceptable to prove an identity by transforming it in a reversible way. With this case, in multiplying-through by a non-zero expression ($cos^2x$), Yves reduced the identity to an utter triviality; to recapture the original identity, one could simply reverse the process and divide the triviality by the same non-zero expression ... even so, the essence of the identity is already established in the transformed version. (Note: We may assume $cos^2 x$ is non-zero, since the appearance of $tan x$ restricts the domain of consideration appropriately.)
$endgroup$
– Blue
Mar 22 at 9:28
1
$begingroup$
@James: if $cne0$, $a=biff ca=cb$. This is what you are missing. You don't have the correct experience in proving identities (no offense but I urge you to review your understanding of equivalence of identities), and you disregarded my comment that explained how to reshape the proof to your taste.
$endgroup$
– Yves Daoust
Mar 22 at 9:34
|
show 7 more comments
$begingroup$
I do not believe your work is correct.
$endgroup$
– James
Mar 22 at 8:41
3
$begingroup$
@James I strongly believe his work is correct and elegant.
$endgroup$
– Le Anh Dung
Mar 22 at 8:43
1
$begingroup$
@James: my approach is fully correct. Sorry to say, you don't understand the principle of proving identities. You can rewrite keeping the denominators, but this is unnecessary. By the way, we can multiply by $cos^2x$ as this factor is guaranteed to be nonzero.
$endgroup$
– Yves Daoust
Mar 22 at 8:48
2
$begingroup$
@James: It is perfectly acceptable to prove an identity by transforming it in a reversible way. With this case, in multiplying-through by a non-zero expression ($cos^2x$), Yves reduced the identity to an utter triviality; to recapture the original identity, one could simply reverse the process and divide the triviality by the same non-zero expression ... even so, the essence of the identity is already established in the transformed version. (Note: We may assume $cos^2 x$ is non-zero, since the appearance of $tan x$ restricts the domain of consideration appropriately.)
$endgroup$
– Blue
Mar 22 at 9:28
1
$begingroup$
@James: if $cne0$, $a=biff ca=cb$. This is what you are missing. You don't have the correct experience in proving identities (no offense but I urge you to review your understanding of equivalence of identities), and you disregarded my comment that explained how to reshape the proof to your taste.
$endgroup$
– Yves Daoust
Mar 22 at 9:34
$begingroup$
I do not believe your work is correct.
$endgroup$
– James
Mar 22 at 8:41
$begingroup$
I do not believe your work is correct.
$endgroup$
– James
Mar 22 at 8:41
3
3
$begingroup$
@James I strongly believe his work is correct and elegant.
$endgroup$
– Le Anh Dung
Mar 22 at 8:43
$begingroup$
@James I strongly believe his work is correct and elegant.
$endgroup$
– Le Anh Dung
Mar 22 at 8:43
1
1
$begingroup$
@James: my approach is fully correct. Sorry to say, you don't understand the principle of proving identities. You can rewrite keeping the denominators, but this is unnecessary. By the way, we can multiply by $cos^2x$ as this factor is guaranteed to be nonzero.
$endgroup$
– Yves Daoust
Mar 22 at 8:48
$begingroup$
@James: my approach is fully correct. Sorry to say, you don't understand the principle of proving identities. You can rewrite keeping the denominators, but this is unnecessary. By the way, we can multiply by $cos^2x$ as this factor is guaranteed to be nonzero.
$endgroup$
– Yves Daoust
Mar 22 at 8:48
2
2
$begingroup$
@James: It is perfectly acceptable to prove an identity by transforming it in a reversible way. With this case, in multiplying-through by a non-zero expression ($cos^2x$), Yves reduced the identity to an utter triviality; to recapture the original identity, one could simply reverse the process and divide the triviality by the same non-zero expression ... even so, the essence of the identity is already established in the transformed version. (Note: We may assume $cos^2 x$ is non-zero, since the appearance of $tan x$ restricts the domain of consideration appropriately.)
$endgroup$
– Blue
Mar 22 at 9:28
$begingroup$
@James: It is perfectly acceptable to prove an identity by transforming it in a reversible way. With this case, in multiplying-through by a non-zero expression ($cos^2x$), Yves reduced the identity to an utter triviality; to recapture the original identity, one could simply reverse the process and divide the triviality by the same non-zero expression ... even so, the essence of the identity is already established in the transformed version. (Note: We may assume $cos^2 x$ is non-zero, since the appearance of $tan x$ restricts the domain of consideration appropriately.)
$endgroup$
– Blue
Mar 22 at 9:28
1
1
$begingroup$
@James: if $cne0$, $a=biff ca=cb$. This is what you are missing. You don't have the correct experience in proving identities (no offense but I urge you to review your understanding of equivalence of identities), and you disregarded my comment that explained how to reshape the proof to your taste.
$endgroup$
– Yves Daoust
Mar 22 at 9:34
$begingroup$
@James: if $cne0$, $a=biff ca=cb$. This is what you are missing. You don't have the correct experience in proving identities (no offense but I urge you to review your understanding of equivalence of identities), and you disregarded my comment that explained how to reshape the proof to your taste.
$endgroup$
– Yves Daoust
Mar 22 at 9:34
|
show 7 more comments
$begingroup$
Expanding gives $sin^2x-tan^2x+cos^2x-1$. The undesired terms cancel because $sin^2x+cos^2x=1$.
answered Mar 22 at 8:29
J.G.J.G.
32.9k23250
$endgroup$
add a comment |
$begingroup$
Expanding gives $sin^2x-tan^2x+cos^2x-1$. The undesired terms cancel because $sin^2x+cos^2x=1$.
answered Mar 22 at 8:29
J.G.J.G.
32.9k23250
$endgroup$
add a comment |
$begingroup$
Expanding gives $sin^2x-tan^2x+cos^2x-1$. The undesired terms cancel because $sin^2x+cos^2x=1$.
answered Mar 22 at 8:29
J.G.J.G.
32.9k23250
$endgroup$
Expanding gives $sin^2x-tan^2x+cos^2x-1$. The undesired terms cancel because $sin^2x+cos^2x=1$.
answered Mar 22 at 8:29
J.G.J.G.
32.9k23250
answered Mar 22 at 8:29
J.G.J.G.
32.9k23250
answered Mar 22 at 8:29
J.G.J.G.
32.9k23250
answered Mar 22 at 8:29
J.G.J.G.
32.9k23250
32.9k23250
add a comment |
add a comment |
$begingroup$
Recall:
$$beginalign
tan^2(x) - sec^2(x) &= -1 \
sin^2(x) + cos^2(x) &= 1
endalign$$
Thus,
$$beginalign
tan^2(x) + 1 &= sec^2(x)\
cos^2(x) - 1 &= -sin^2(x)
endalign$$
We also note that $cos(x)$ is an even function, and thus $cos(-x) = cos(x)$. Thus, the formula becomes:
$$(tan^2(x) + 1)(cos^2(-x) - 1) = -sec^2(x)sin^2(x) = - fracsin^2(x)cos^2(x) = -tan^2(x)$$
answered Mar 22 at 8:30
Eevee TrainerEevee Trainer
10.1k31742
$endgroup$
add a comment |
$begingroup$
Recall:
$$beginalign
tan^2(x) - sec^2(x) &= -1 \
sin^2(x) + cos^2(x) &= 1
endalign$$
Thus,
$$beginalign
tan^2(x) + 1 &= sec^2(x)\
cos^2(x) - 1 &= -sin^2(x)
endalign$$
We also note that $cos(x)$ is an even function, and thus $cos(-x) = cos(x)$. Thus, the formula becomes:
$$(tan^2(x) + 1)(cos^2(-x) - 1) = -sec^2(x)sin^2(x) = - fracsin^2(x)cos^2(x) = -tan^2(x)$$
answered Mar 22 at 8:30
Eevee TrainerEevee Trainer
10.1k31742
$endgroup$
add a comment |
$begingroup$
Recall:
$$beginalign
tan^2(x) - sec^2(x) &= -1 \
sin^2(x) + cos^2(x) &= 1
endalign$$
Thus,
$$beginalign
tan^2(x) + 1 &= sec^2(x)\
cos^2(x) - 1 &= -sin^2(x)
endalign$$
We also note that $cos(x)$ is an even function, and thus $cos(-x) = cos(x)$. Thus, the formula becomes:
$$(tan^2(x) + 1)(cos^2(-x) - 1) = -sec^2(x)sin^2(x) = - fracsin^2(x)cos^2(x) = -tan^2(x)$$
answered Mar 22 at 8:30
Eevee TrainerEevee Trainer
10.1k31742
$endgroup$
Recall:
$$beginalign
tan^2(x) - sec^2(x) &= -1 \
sin^2(x) + cos^2(x) &= 1
endalign$$
Thus,
$$beginalign
tan^2(x) + 1 &= sec^2(x)\
cos^2(x) - 1 &= -sin^2(x)
endalign$$
We also note that $cos(x)$ is an even function, and thus $cos(-x) = cos(x)$. Thus, the formula becomes:
$$(tan^2(x) + 1)(cos^2(-x) - 1) = -sec^2(x)sin^2(x) = - fracsin^2(x)cos^2(x) = -tan^2(x)$$
answered Mar 22 at 8:30
Eevee TrainerEevee Trainer
10.1k31742
answered Mar 22 at 8:30
Eevee TrainerEevee Trainer
10.1k31742
answered Mar 22 at 8:30
Eevee TrainerEevee Trainer
10.1k31742
answered Mar 22 at 8:30
Eevee TrainerEevee Trainer
10.1k31742
10.1k31742
add a comment |
add a comment |
$begingroup$
- $1+tan^2x=sec^2x$
- $sin^2x+cos^2x=1$
$cos x=cos(-x)$ i.e. $cos x$ is an even function.- $sec x=1/cos x$
$$underbraceleft(1+tan^2xright)_=sec^2xunderbraceleft(cos^2(-x)-1right)_cos xtext is even function=-sec^2xcdotsin^2x=-tan^2x $$
answered Mar 22 at 8:31
Paras KhoslaParas Khosla
2,883523
$endgroup$
add a comment |
$begingroup$
- $1+tan^2x=sec^2x$
- $sin^2x+cos^2x=1$
$cos x=cos(-x)$ i.e. $cos x$ is an even function.- $sec x=1/cos x$
$$underbraceleft(1+tan^2xright)_=sec^2xunderbraceleft(cos^2(-x)-1right)_cos xtext is even function=-sec^2xcdotsin^2x=-tan^2x $$
answered Mar 22 at 8:31
Paras KhoslaParas Khosla
2,883523
$endgroup$
add a comment |
$begingroup$
- $1+tan^2x=sec^2x$
- $sin^2x+cos^2x=1$
$cos x=cos(-x)$ i.e. $cos x$ is an even function.- $sec x=1/cos x$
$$underbraceleft(1+tan^2xright)_=sec^2xunderbraceleft(cos^2(-x)-1right)_cos xtext is even function=-sec^2xcdotsin^2x=-tan^2x $$
answered Mar 22 at 8:31
Paras KhoslaParas Khosla
2,883523
$endgroup$
- $1+tan^2x=sec^2x$
- $sin^2x+cos^2x=1$
$cos x=cos(-x)$ i.e. $cos x$ is an even function.- $sec x=1/cos x$
$$underbraceleft(1+tan^2xright)_=sec^2xunderbraceleft(cos^2(-x)-1right)_cos xtext is even function=-sec^2xcdotsin^2x=-tan^2x $$
answered Mar 22 at 8:31
Paras KhoslaParas Khosla
2,883523
answered Mar 22 at 8:31
Paras KhoslaParas Khosla
2,883523
answered Mar 22 at 8:31
Paras KhoslaParas Khosla
2,883523
answered Mar 22 at 8:31
Paras KhoslaParas Khosla
2,883523
2,883523
add a comment |
add a comment |
$begingroup$
We know
$(tan x)' = 1 + tan^2 x = frac1cos^2 x$ and- $cos (-x) = cos x$
Now, it follows immediately
begineqnarray* (tan^2(x)+1)(cos^2(-x)-1)
& = & (tan x)'left(frac1(tan x)'-1right) \
& = & 1 - (tan x)' \
& = & 1 - (1 + tan ^2 x) \
& = & - tan ^2 x \
endeqnarray*
answered Mar 22 at 9:11
trancelocationtrancelocation
13.6k1829
$endgroup$
add a comment |
$begingroup$
We know
$(tan x)' = 1 + tan^2 x = frac1cos^2 x$ and- $cos (-x) = cos x$
Now, it follows immediately
begineqnarray* (tan^2(x)+1)(cos^2(-x)-1)
& = & (tan x)'left(frac1(tan x)'-1right) \
& = & 1 - (tan x)' \
& = & 1 - (1 + tan ^2 x) \
& = & - tan ^2 x \
endeqnarray*
answered Mar 22 at 9:11
trancelocationtrancelocation
13.6k1829
$endgroup$
add a comment |
$begingroup$
We know
$(tan x)' = 1 + tan^2 x = frac1cos^2 x$ and- $cos (-x) = cos x$
Now, it follows immediately
begineqnarray* (tan^2(x)+1)(cos^2(-x)-1)
& = & (tan x)'left(frac1(tan x)'-1right) \
& = & 1 - (tan x)' \
& = & 1 - (1 + tan ^2 x) \
& = & - tan ^2 x \
endeqnarray*
answered Mar 22 at 9:11
trancelocationtrancelocation
13.6k1829
$endgroup$
We know
$(tan x)' = 1 + tan^2 x = frac1cos^2 x$ and- $cos (-x) = cos x$
Now, it follows immediately
begineqnarray* (tan^2(x)+1)(cos^2(-x)-1)
& = & (tan x)'left(frac1(tan x)'-1right) \
& = & 1 - (tan x)' \
& = & 1 - (1 + tan ^2 x) \
& = & - tan ^2 x \
endeqnarray*
answered Mar 22 at 9:11
trancelocationtrancelocation
13.6k1829
answered Mar 22 at 9:11
trancelocationtrancelocation
13.6k1829
answered Mar 22 at 9:11
trancelocationtrancelocation
13.6k1829
answered Mar 22 at 9:11
trancelocationtrancelocation
13.6k1829
13.6k1829
add a comment |
add a comment |
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$begingroup$
$tan^2(x)+1=sec^2(x)=frac 1 cos^2(x)$ and $1-cos^2(-x)=1-cos^2(x)=sin^2(x)$.
$endgroup$
– Kavi Rama Murthy
Mar 22 at 8:25
1
$begingroup$
I wonder how you failed to solve this using the Pythagorean theorem, this is straightforward.
$endgroup$
– Yves Daoust
Mar 22 at 8:32