Use the identity tan(x) = sin(x) / cos(x) in the left hand side of the given identity.
tan2(x) - sin2(x) = sin2(x) / cos2(x) - sin2(x)
= [ sin2(x) - cos2(x) sin2(x) ] / cos2(x)
= sin2(x) [ 1 - cos2(x) ] / cos2(x)
= sin2(x) sin2(x) / cos2(x)
= sin2(x) tan2(x) which is equal to the right hand side of the given identity.
Answers
csc(x)=
sin(x)
1
\sin(x) = \dfrac{1}{\csc(x)}sin(x)=
csc(x)
1
\sec(x) = \dfrac{1}{\cos(x)}sec(x)=
cos(x)
1
\cos(x) = \dfrac{1}{\sec(x)}cos(x)=
sec(x)
1
\cot(x) = \dfrac{1}{\tan(x)} = \dfrac{\cos(x)}{\sin(x)}cot(x)=
tan(x)
1
=
sin(x)
cos(x)
\tan(x) = \dfrac{1}{\cot(x)} = \dfrac{\sin(x)}{\cos(x)}tan(x)=
cot(x)
1
=
cos(x)
sin(x)
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Notice how a "co-(something)" trig ratio is always the reciprocal of some "non-co" ratio. You can use this fact to help you keep straight that cosecant goes with sine and secant goes with cosine.
The following (particularly the first of the three below) are called "Pythagorean" identities.
sin2(t) + cos2(t) = 1
tan2(t) + 1 = sec2(t)
1 + cot2(t) = csc2(t)
Note that the three identities above all involve squaring and the number 1. You can see the Pythagorean-Thereom relationship clearly if you consider the unit circle, where the angle is t, the "opposite" side is sin(t) = y, the "adjacent" side is cos(t) = x, and the hypotenuse is 1.
We have additional identities related to the functional status of the trig ratios:
sin(–t) = –sin(t)
cos(–t) = cos(t)
tan(–t) = –tan(t)
Notice in particular that sine and tangent are odd functions, being symmetric about the origin, while cosine is an even function, being symmetric about the y-axis. The fact that you can take the argument's "minus" sign outside (for sine and tangent) or eliminate it entirely (for cosine) can be helpful when working with complicated expressions.
Angle-Sum and -Difference Identities
sin(α + β) = sin(α) cos(β) + cos(α) sin(β)
sin(α – β) = sin(α) cos(β) – cos(α) sin(β)
cos(α + β) = cos(α) cos(β) – sin(α) sin(β)
cos(α – β) = cos(α) cos(β) + sin(α) sin(β)
\tan(\alpha + \beta) = \dfrac{\tan(\alpha) + \tan(\beta)}{1 - \tan(\alpha) \tan(\beta)}tan(α+β)=
1−tan(α)tan(β)
tan(α)+tan(β)
\tan(\alpha - \beta) = \dfrac{\tan(\alpha) - \tan(\beta)}{1 + \tan(\alpha) \tan(\beta)}tan(α−β)=
1+tan(α)tan(β)
tan(α)−tan(β)
By the way, in the above identities, the angles are denoted by Greek letters. The a-type letter, "α", is called "alpha", which is pronounced "AL-fuh". The b-type letter, "β", is called "beta", which is pronounced "BAY-tuh".
Content Continues Below
Double-Angle Identities
sin(2x) = 2 sin(x) cos(x)
cos(2x) = cos2(x) – sin2(x) = 1 – 2 sin2(x) = 2 cos2(x) – 1
\tan(2x) = \dfrac{2 \tan(x)}{1 - \tan^2(x)}tan(2x)=
1−tan
2
(x)
2tan(x)
Half-Angle Identities
\sin\left(\dfrac{x}{2}\right) = \pm \sqrt{\dfrac{1 - \cos(x)}{2}}sin(
2
x
)=±
2
1−cos(x)
\cos\left(\dfrac{x}{2}\right) = \pm \sqrt{\dfrac{1 + \cos(x)}{2}}cos(
2
x
)=±
2
1+cos(x)
\tan\left(\dfrac{x}{2}\right) = \pm \sqrt{\dfrac{1 - \cos(x)}{1 + \cos(x)}}tan(
2
x
)=±
1+cos(x)
1−cos(x)
= \dfrac{1 - \cos(x)}{\sin(x)}=
sin(x)
1−cos(x)
= \dfrac{\sin(x)}{1 + \cos(x)}=
1+cos(x)
sin(x)
The above identities can be re-stated by squaring each side and doubling all of the angle measures. The results are as follows:
\sin^2(x) = \frac{1}{2} \big[1 - \cos(2x)\big]sin
2
(x)=
2
1
[1−cos(2x)]
\cos^2(x) = \frac{1}{2} \big[1 + \cos(2x)\big]cos
2
(x)=
2
1
[1+cos(2x)]
\tan^2(x) = \dfrac{1 - \cos(2x)}{1 + \cos(2x)}tan
2
(x)=
1+cos(2x)
1−cos(2x)
Sum Identities
\sin(x) + \sin(y) = 2 \sin\left(\dfrac{x + y}{2}\right) \cos\left(\dfrac{x - y}{2}\right)sin(x)+sin(y)=2sin(
2
x+y
)cos(
2
x−y
)
\sin(x) - \sin(y) = 2 \cos\left(\dfrac{x + y}{2}\right) \sin\left(\dfrac{x - y}{2}\right)sin(x)−sin(y)=2cos(
2
x+y
)sin(
2
x−y
)
\cos(x) + \cos(y) = 2 \cos\left(\dfrac{x + y}{2}\right) \cos\left(\dfrac{x - y}{2}\right)cos(x)+cos(y)=2cos(
2
x+y
)cos(
2
x−y
)
\cos(x) - \cos(y) = -2 \sin\left(\dfrac{x + y}{2}\right) \sin\left(\dfrac{x - y}{2}\right)cos(x)−cos(y)=−2sin(
2
x+y
)sin(
2
x−y
)
Product Identities
\sin(x) \cos(y) = \frac{1}{2} \big[\sin(x + y) + \sin(x - y)\big]sin(x)cos(y)=
2
1
[sin(x+y)+sin(x−y)]
\cos(x) \sin(y) = \frac{1}{2} \big[\sin(x + y) - \sin(x - y)\big]cos(x)sin(y)=
2
1
[sin(x+y)−sin(x−y)]
\cos(x) \cos(y) = \frac{1}{2} \big[\cos(x - y) + \cos(x + y)\big]cos(x)cos(y)=
2
1
[cos(x−y)+cos(x+y)]
\sin(x) \sin(y) = \frac{1}{2} \big[\cos(x - y) - \cos(x + y)\big]sin(x)sin(y)=
2
1
[cos(x−y)−cos(x+y)]
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You will be using all of these identities, or nearly so, for proving
Hope this helps
Answer:
विद्यालय में रग्क संगीत सम्मेलन करने की अनुमति लेने हेतु अपने प्रधानाचार्य से अनुरोध की जिए।
Explanation: